today we're going to be looking at the trig addition formula okay we're going to be able to use the double angle formula and the trig addition formula and right expressions in the form a cos theta plus or minus b sine theta in the form of r cos theta plus alpha okay so so and the addition formula allows us to deal with a sum or a difference for example 60 degrees could be written as 20 degrees and plus 40 degrees or 60 degrees could be written as 90 degrees minus 30 degrees okay we're manipulating angles here and the reasons why will become apparent in a moment okay but the basics of it is is that we are going to write angles as either a product sorry not a product as a sum or a difference okay and this will help us to manipulate and find the answer so these are the formulas you need okay they are in the formula booklet but obviously it's easier if you just know them okay that's why today we're going to have them in front of us all the time because but what you will find is that you don't want to memorize them but then you end up memorizing them they're in your head okay so um this is what we're going to be looking at today okay so do you notice with sine the answer is sine cos cosine okay so with psi it's always sine cos and then it's the same sign plus and then it's cosine okay and it's a b a b okay with the minus it's just sine cos minus cosine with cos though it's always cos cos sine sine and then it's the opposite if that says plus you have to minus here sine sine and then if it's minus cos cos plus sine sine and then with tan okay it's tan plus tan over one minus tan tan and then when it's the minus the size switch so it's tan minus tan over one plus tan tan okay right so there is a proof of um it's not needed for the exam which isn't needed for so how could we prove the result that sine of a plus b equals sine a cos no sorry given that sine of a plus b equals sine a cos b plus cos a sine b to prove the identity for sine a minus b so what we're going to do is instead of writing b we're going to substitute in minus b so this is the same as saying sine of a plus minus b which equals sine of a cos minus b plus and then we've got sine sorry cos a and then we're going to have sine minus b so what's going to happen now we know does anybody know a fun fact about cos of minus b no we cannot we could but if we think about the graph cos it's a reflection okay it's a reflection in the y axis so cos of minus 10 is the same of cos 10. so cos of minus b equals cos of b so we have sine a cos of minus b is just cos of b because cos graph is a reflection in the y axis and then cos a sine minus b we can take that minus out to have minus cos a sine b and that's how we've shown that's how we've proved given this result sine of a minus b okay this has come up in an exam before and where people have got tripped up is the cos of minus b now cos of minus b is the same as cos of b because the cos graph is a reflection in the y axis when in doubt draw it out okay right how about cos given that we have sine of a minus b how could we prove cos of a plus b we know that cos of x equals sine lots of pi minus two over x in this case our x is a plus b so we're going to substitute in minus a plus b okay so this is the same as writing sine of pi over 2 minus a plus b now i can rewrite this as sine pi of 2 minus a minus b but what i want to do is group two of them together okay now using what we just said that pi minus pi over 2 minus a number is cos of that number i'm going to group pi over 2 minus a number here so i'm going to have sine pi over 2 minus a minus b okay so this pi over 2 minus a is now going to be my a and this minus b is now going to be my b okay so using the sine rule up here because this is all we've been given and we have um these two here so that should actually say sorry everyone i just give myself a bit more room plus minus b remember because we've only been given this top formula here so we have to use this one we're now going to have sine of a and our a is pi over 2 minus a cos of b which is cos of minus b plus cos of a which is pi over 2 minus a sine b which is sine of minus b now sine pi over 2 minus a we know is actually in fact cos a cos of minus b is in fact just cos b cos of pi over 2 minus a is in fact sine of a sine of minus b i can bring that minus b at the minus out here and then have sine b so then that's how we've shown what cos of a plus b equals okay first things first what is tan sine over cos so tan of a plus b equals sine of a plus b over cos of a plus b okay next step i know what sine of a plus b is and i know what cos of a plus b is and it is said in the question that i can use this so this equals sine of a plus b is sine a cos b plus cos a sine b all of that is over um cos a cos b minus sine a sine b so you're now going to divide every term by cos a cos b and remember when we're manipulating functions as long as we're ensuring that we do the same to every variable we're okay so now you can see here sine a over cos a is tan a and cos b over cos b is one so this is how we get tan a here then for this term cos a over cos a is one sine b over cos b is in fact tan b so plus tan b then cos a cos b over cos a cos b is one sine a over cos a is tan a sine b over cos b is tan b so that's tan a tan b so given that 2 sine x plus y equals 3 cos x minus y express tan x in terms of tan y so it's asking us for tan x in terms of tan y so what that means is we need to have tan x equals so step one i'm going to be using the addition formula to rewrite sine and cos okay so when we are proving and expressing in terms of time remember there's more than one way some are quicker than others um but you should always get to the same answer okay so sine of x plus sine b now if you've got your book in front of you we're going to have that's going to be 2 lots of sine x cos y plus cos x sine y equals 3 cos x cos y plus um oh make sure this goes in brackets um sine x sine y perfect now i can expand my brackets so i'm going to get 2 sine x cos y plus 2 cos x sine y equals 3 cos x cos y plus 3 sine x sine y some of you might do that in the first step that's completely fine now i want tan the only way i'm going to get tan is to divide okay and if i think here if i've got a sine x how could i get tan i need to divide by cos so if i divide this by cos x here i also this cos y i want to get rid of it i want it to be a one so i'm going to divide by cos x cos y so if i divide everything by cos x cos y okay and whatever i do to one term i have to do to them all so i'm then going to end up with um sine over cos that's going to give me tan cos over cos is one so i'm going to get 2 tan x plus and then here i'm going to have 2 because x over cos x is 1 sine y over cos y is tan y to tan y equals cos x over cos x and cos y because y is one so that's just three and sine x over cos x is tan x and sine y over cos y is tan y so this is where i'm at but i want it to have tan x equals okay so whenever we rearrange and we've got more than one um shown i need to so i'm gonna put all my tan x's on one side everything else on the other side so simple rearranging so 2 tan x minus 3 tan x 10 y equals 3 minus 210 y is everyone happy how i got that okay and then now because i want tan x on its own i'm going to factorize tan x on the left hand side so tan of x is 2 minus 3 tan y which equals 3 minus 2 tan y and then how do i get tan of x on its own i'm gonna divide both sides so tan of x equals three minus two tan y over two minus three tan y okay again this is very similar to rearranging at gcse level but instead of x we have a trig function and we just have to rearrange it at the beginning but this is just manipulating algebra that's all we're doing manipulating so exercise 3a please can you go to exercise 3a and no i don't think it is exercise for you i think it's exercise 4a and can you please have a look at um okay can you do question [Music] two [Music] question five question seven [Music] question nine a c d so what we should really know okay and in the uk if you were doing your a levels there and this wasn't an international you'd have to know this memorize this okay and they actually memorized this for the gcse in the uk so a few what you need to know so you need to know zero 30 degrees 45 degrees 60 degrees 90 degrees okay this is what we need to know and again you should know that if we converted this to pi okay we know that 90 is pi over two so 45 is pi over four zero is zero um sixty and thirty is going to be pi over 3 and pi over 6. so again if we're dealing with radians or you know we should be able to know this okay and what we're going to have is we're going to um we're going to find out what sine of x is cos of x and tan of x okay let's see if we what we can do without using our calculator first of all when sine of x equals zero if you think about the graph when sine of x equals zero what does y equal zero amazing what else do we know about sine of x when yes sine of x has a maximum when x equals 90 and y equals one perfect okay so we know that how about with cos does anybody know with cos if we look at the graph cos of x equals zero what is our y coordinate when cos of x equals zero it is one absolutely okay and when cos of x equals 90 zero absolutely when tan of x equals zero think about the tang graph when tan of x equals zero yeah it is zero and when tan of x equals 90 yes because we know it tends to infinity so okay perfect so we know those which is amazing what we need to try and get used to is knowing the rest so we can fill these in so if we did sine of 30 we're going to get a half sine of 45 we're going to get root 2 over 2 and sine of 16 is in fact root 3 over 2. now because we know that cos is a translation of sine okay um it's reversed so we have a half root 2 over 2 root 3 over 2. then 10 slightly different so tan of 30 is root 3 over 3. tan of 45 is in fact 1 and tan of 16. is root 3. using a suitable angle formula show that sine of 15 degrees equals root 6 minus root 2 over 4. okay so we have a few key numbers that we like to work with okay we can work with 0 30 45 16 and 95 oh sorry not 95 and 90. is there a way we can make 15 absolutely 45 minus 30 okay so this is going to be sine of 15 i can rewrite that as sine of 45 minus 30. okay we have a double angle formula we have a look on in our books because we haven't memorized these year and this is actually going to be sine remember our a is 45 and our b is 30. so we're going to have sine of 45 minus 30 is rewritten as sine of a which is 45 cos of b which is 30 minus sorry cos of a which is 45 sine of b which is 30 and again we're just using the formula to fill it in aren't we okay all we're doing is substitution nice and easy now using that picture you've just taken on your phone okay you know that sine of 45 is root 2 over 2 cos of 30 is root 3 over 2 minus cos of 45 which is root 2 over 2 and sine of 30 which is a half so all i've done now is written the exact values of sine or cos okay and now i'm going to multiply expand you know and take away so here we have a fraction problem now so root 2 times root 3 is in fact root six over four minus root two times one is root two over four lucky for us they have the same denominator so we've not gonna multiply by anything so we can just write root six minus root two over four therefore shown okay so um second question given that sine of a equals minus three over five and 180 degrees is less than a is less than 270 and that cos b equals minus 12 over 13. find the value a of cos of a minus b and b tan of a plus b okay so step one let's do part a first it's told us cos a minus b okay so cos of a minus b equals and let's write it down what does it equal cos a cos b plus sine a sine b okay so we've got that written down now it tells us that cos b equals and we have a sine a okay well hold on a minute i know that cos b is minus 12 over 13. so everywhere it says cos b i'm going to write minus 12 over 13. but i have sine a great i don't have a sine b yet and i don't have a cos a but that's okay because i know a formula that can help us okay i know that sine squared plus cos squared equals one i'm going to use that to then find the opposite so i know then that um if sine squared of a plus cos squared a equals 1 okay that means that cos of a equals the square root of 1 minus sine squared a and i could write that if sine squared b plus cos squared b equals 1 that means sine b equals the square root of 1 minus cos squared b now i have been given sine a which is minus 3 over 5 so cos a so now i'm going to have cos a equals square root of 1 minus minus 3 over 5 squared because that's what sine of a equals minus 3 over 5. okay i'm going to square it so that means that cos a is 1 minus 9 over 25 which is 16 over 25 square rooted which is 4 over 5 plus or minus 4 over 5 when we do the square root okay and then we get plus or minus four over five but this is where we have to think about the graph again we always refer to the graph it says 180 degrees to 270 degrees okay if we draw the cross graph okay we know this is 180 this is in fact 270. okay so it's got to be in between here so do we want a positive or a negative four over five we have a look the graph is in between here so we can see the graph is here use a different color the graph is here is where we want it and this is negative so we want to have the negative 4 over 5. we're going to reject the positive 4 over 5. so we're going to reject the positive and we're in fact only going to keep the negative so that means that cos of a in fact just equals negative 4 over 5 because of the interval that it's in is in between 180 and 270. okay so now we figured out cos of a perfect now we're going to do the same thing and find sine b so sine of b equals the square root of 1 minus cos squared of b now it tells us that cos of b is minus 12 over 13 so it's minus 12 over 13 squared okay so we're going to get sine of b equals 1 minus 144 over 169 which is 25 over 16 169 25 over 169 square rooted which is in fact plus or minus 5 over 13. okay again use your calculator for that right now let's think of the sine graph i think the sine graph is here this is 180 here is 270 okay and if we look at the graph between 118 and 270 is a negative value okay so we're going to take the negative value so we're going to have negative so sine b equals negative 5 over 13. i'm so sorry this is so messy okay right now we have our rule and we have every single variable now so now we can substitute it so we know that cos of a minus b okay equals cos a which it told us in no we just figured it out is minus four over five actually should i just let's highlight everything okay sine of a is in the question cos b's in the question cos a is in the question we figured out and sine b we figured out okay so there we are i've highlighted them so we can see them okay so it's cos a which is minus four over five multiplied by cos b which is in the question minus 12 over 13. then add sine of a which is in the question minus 3 over 5 multiplied by sine of b which we just figured out minus 5 over 13. okay so then what's going to happen is um cos of i minus b equals 12. i'm going to get 48 over 65 plus 15 over 65 which equals um 63 over 65 so this is the value of cos of a minus b it's 63 over 65 within the interval of 180 and 270. okay i'd just like to point out in the book the textbook it actually says that um the angle of uh b is obtuse if b is obtuse it means it's in the interval of 90 and 180 which when we're figuring out sine we would in fact take the positive value okay what i've done um in this question is i've used the same interval that b is between 180 and 270 which gives us the negative 5 over 13. so for the question that we've answered this is completely correct so for part and b we're going to stick with the same interval again 180 degrees is less than b which is less than 270. so for this question that's what we've done okay so again we're saying that 180 degrees is less than b was less than 270 degrees so using that same interval we already figured out what um the missing values were so sine b and cos a so we always need to write down that tan of a plus b and using our formula we know that that's equal to tan of a plus tan of b all over 1 minus tan a tan b so we already know that sine of a equals minus 3 over 5. we know that sine of b equals minus 12 no sorry minus 5 over 13. we know that cos of a equals negative 4 over 5 and cos of b equals negative 12 over 13. that was from earlier and two of them we got in question we know that tan is silly cow so tan of a is going to equal sine of a over cos of a so tan of a is going to equal minus 3 over 5 over cos of a which is minus four over five which just equals three over four we know that tan of b is sine b over cosby which is um minus 5 over 13 over minus 5 over minus 12 over 13 which is in fact 5 over 12. now we have those values we can substitute them in so that means then that tan of a plus b equals tan of a which is three-quarters plus tan of b which is 5 12 all over one minus tan of a three-quarters multiplied by tan of b which is five over twelve okay and then if we figure that out 3 over 4 in terms of 12 is 9 over 12 plus 5 over 12 is 14 over 12. all of that over 3 times 1 minus 15 over 48 which equals 14 over 12 over 33 over 48 did anyone get an answer 8 over 9 which equals eight over nine yeah no um 50 6 over 33 so that's what tan of a plus b equals 56 over 33. all right so test your understanding now remember we like to use the numbers 0 30 45 60 and 90. okay so cos of 75 how can i make 75 30 and 45 so 30 plus 45 so i know i'm going to have to do cos of 30 plus 45 so this is going to be my cos of a plus b we know that cos of a plus b is cos of a so cos a cos b minus sine a sine b so this is going to equal cos 30 cos 45 minus sine 30 sine 45 now we know what these are okay and if you don't know what they are you can put them into your calculator so it's 1 over root 2 times root 3 over 2 minus 1 over root 2 times a half which i think we did this example earlier root six minus root two over four perfect and then tan of 75 um again we're going to use the same so we're gonna have tan of 30 plus 45 which is going to give me tan of 30 plus tan of 45 over 1 minus 10 30 tan 45. now of course you could put this into your calculator but you're not going to get the marks okay because it's it's one of those show that questions now we could do this way another way you could do it is you know that tan of 75 is also sine of 75 over cos of 75 okay so you could use the sine and then divide that by the cos either way is fine um so then what you're going to have we know that tan of 30 is um root 3 over 3 plus 1 divided by 1 minus root 3 over 3 times 1 which is in fact going to give me when i put that in my calculator two plus root three again you could have done this a slightly different way which i'll show you on the next page actually here we go so on the next page they've done sine of 75 over cos of 75 okay because that we already figured out cos of 75 so you can already use that and you just use sine of 75 and then divide the two values together to get the answer okay either way it's the same okay so now i have a bit of a challenge for you to do once you've done the challenge um you are then going to move on to some of the questions in the textbook so this was a question in 2013 so it says given that 2 cos of x plus 50 equals sine of x plus 40. so step one write cos of x plus 50 and sine of x plus 40 as the expanded addition formula so we know that we're going to have two lots if this is my a and b okay cos of a plus b is cos x cos 50 um minus 2 sine x sine 50 okay so that's what the cos equals and that actually equals sine of x plus 40. that's that's sine of a plus b so i'm going to get sine a which is x cos b which is 40 plus cos a which is x sine b which was 40. okay so that equals that okay how um what am i now going to do okay what do we know again if we have a look we have a cos 50 a sine 50 okay now what do we know about the relationship between sine and cos okay we know that sine of x equals cos of 90 minus x and we know that cos of x equals sine of 90 minus x okay because we know it's a translation don't we okay so this is this is quite challenging here because actually so that means that sine of 50 is in fact cos of 90 minus 50 which is cos of 40 and cos of 40 at 50 equals cos of 90 minus 50 which is sorry sine of 90 minus 50 which is sine of 40. and the reason why i've done that is because we have tan 40. okay so we need to make sure that we then rewrite everything that we've just learned so we're going to have 2 cos x cos 50 is now in fact sine 40 minus 2 sine x sine 50 is in fact cos 40 equals sine x cos 40 is in fact sine nope we don't know we already got 40. so that's cos 40 plus cos x sine 40. okay so now we've got to divide by um something to make sure we get tan okay and the only thing we could divide by is going to be cos x cos 40. so i'm going to divide everything by cos x cos 40. yeah that'll give me 10. cos x cos 40 that'll give me 10 yep cos x cos 40. that'll also give me a turn cos x cos 40. brilliant that'll give me tan too so what we're going to have is 2 cos x sine 40 over cos x cos 40 is in fact 2 tan 40 minus 2 sine x cos x cos 40 over cos 40 so that's just 2 tan x equals sine x over cos x and the cos 40 is going to cancel so that's tan x plus cos x sine 40 over cos x cos 40 is just going to be tan 40 so now i need to rearrange so if i move the tan 40s to the left hand side i'm going to have tan 40 and all the 10x is to the right hand side is three tan x okay i'm going to rewrite that because in our question they've got tan x on the left hand side i've turned 40 and then to find one tan x i'm going to divide by 3 so tan x equals a third of tan 40. okay so you need to do x says 4b question three five six part b and c and question nine part a b and c um the angle in half okay which is really amazing now how they've derived this formula is actually by using um the addition formula but instead of having sine of a plus b they've got sine of a plus a because a plus a is in fact 2a they've substituted it in and then simplified now cos of 2a the identity is cos squared a minus sine squared a but as you know we know that cos squared a plus sine squared a equals one so by sub rearranging this and substituting in we then um managed to make two other identities and then we've tan again instead of writing um a plus b it's a plus a and then when we substitute in we're going to get this formula okay so nice and simple so you can derive these formulas if you need to so what we're going to do is we're going to take a look at a few examples so if we have a look at part a it says use the double angle formula to write each of the following as a single trigonometric ratio so the minute we've got two trigonometric ratios we just want a single so i have cos squared and sine squared so if i look on here it actually matches because of 2x which is cos squared squared x minus sine squared x so we know oh sorry i don't know what happened there so in this case this is going to give us cos of 2 x okay equals cos squared x minus sine squared x so you can see here it matches up our x is in fact 50. so we know that cos of 2 lots of 50 equals cos of 100 which is cos squared of 50 minus 5 squared of 50. so our actual answer is cos of 100 and obviously we're working in degrees then for the next one we have 2 tan of pi over 6 over 1 minus 10 squared pi over 6. if we have a look this matches this one here our x is in fact pi over six so we know that this is going to equal tan of two lots of pi over six which equals tan of pi over 3. then for the last one is slightly different because we have 4 sine of 70 over sec of 70. so we need to do a slight bit of rearranging so we know we have 4 sine of 70 multiplied by we know that sec is one over cos so multiplied by one over oh sorry it's not multiplied at all and this is cause so 4 over sorry 4 sine 70 and then it's divided by sec of 17. so we're going to have 4 sine of 17 divided by 1 over cos of 17 okay which we know stay change flip so 4 sine of 17 multiplied by cos of 17. so this is actually going to simplify to 4 sine 70 cos 70. okay so if we have a look what rule this match to and this matches to this rule here okay sine of 2x equals 2 sine x cos x but here we've got four lots so we need to factorize out a 2 to get 2 lots of 2 sine 70 cos 17. now we know that we could rewrite 2 sine 70 cos 70 as sine of 2 lots of 70 which is sine of 140 so we have 2 lots of sine 140 which we're then going to just write as 2 sine 140 so here are some more examples of how we might use the double angle formula so given that x equals three sine theta and y equals three minus four cos of two theta eliminate theta and express y in terms of x so we're going to need an equation where y equals um something with x in so first of all it's given us x equals 3 sine theta so we could rearrange that and if we divide both sides by three we get sine theta equals x over three okay then for the second one we have y equals three minus four cos of 2 theta now we have a sine here so and we want to eliminate theta so when i'm choosing which double angle formula i need for cos of 2 theta i'm going to look for the one with just a sign in because we've already been given what sine is so i'm going to use 1 minus 2 sine squared theta so i'm going to have y equals 3 minus 4 lots of 1 minus 2 sine squared theta so i'm going to get y equals three minus four minus you know minus four times minus two is plus eight sine squared theta so y equals three minus four which is minus one plus eight sine squared theta now we know that sine theta actually equals x over three and we know that sine squared theta is in fact sine of theta all squared so i'm going to substitute that in so therefore y equals minus 1 plus eight lots of x over three squared so i'm going to just simplify that slightly um so y equals um eight x squared over nine minus one so now we've eliminated theta and we've expressed y in terms of x given that cos x equals three quarters and x is acute find the exact value of sine 2x and tan 2x so what equation do i know that relates cos and sine so that if i have cos how could i find sine forget that it's a double angle formula for a moment i just know that to find tan i need sinu and for sine i need to find cos i do know that cos squared x plus sine squared x equals one so using that to begin with cos squared x plus sine squared x equals one so that means sine of x equals the square root of one minus cos squared x so sine of x equals the square root of one minus we know that cos of x is three quarters so that's going to be three quarters squared so sine of x equals root 1 minus 9 over 16 um which is root seven over four yes it is okay root seven over four but again we have a plus or minus now it tells us that um x is acute so if i think about x being acute for the sine graph that means that we're going to have x is less than 90 which we can see here all positive values so we're going to reject the negative and we're going to use sine of x equals the positive root over 4. okay so now we have what causes we have what sine is we can get back to the actual question so we want to find so part a we want to find sine of 2x now sine of 2x we know it is 2 sine x cos x okay so this equals two sine x cos x which is going to equal two lots of sine x is in fact root seven over four and cos x is in fact three over four so we're gonna have six root seven over 16 which is in actual fact 3 root 7 over 8. so nice and simple now we've got all the information okay part b we need tan 2 of x now when we look at tan 2 of x the formula is 2 tan of x over 1 minus tan squared x so that equals 2 tan x over 1 minus tan squared x so we need tan of x luckily we have sine of x and cos of x so if we just do a little side note so tan of x equals silicon sine over cos sine of x we just worked out is 3 um sorry that was the sine of 2x so root 7 over 4 divided by cos of x which we know is three over four so i'm going to get root seven over three perfect now we've got that we can substitute in so tan two of x is going to equal two lots of tan x which we know is root seven over three so two lots of roots seven eight three over one minus tan squared x which is root 7 over 3 all squared i'm going to substitute that into my calculator and i'm going to get 3 root 7. so nice and easy and that is a few more examples on how we solve equations um sorry we're not solving equations how we write equations using the double angle formula so exercise 4c please can you do questions 6 13 14 15 and 16. moving on to solving trigonomet trigonometric equations so what we're going to do now is we're going to be solving the equations just like we normally would but now that we've learned about the addition formula and the double angle formula we may need to use that first to rearrange the equation into a form that we can then solve so let's have a look at this equation solve three cos two of x minus cos x plus two now at the minute i'm really not sure how i'm gonna solve that so i'm going to need to manipulate this now when i look at this the first thing that pops to my head is i wish this was a quadratic well fear not we can make this into a quadratic because we know that cos 2 of x can be written in terms of cos squared so it's 2 cos squared x minus 1. so i'm going to substitute that in and i'm going to have 3 lots of 2 cos squared x minus 1 minus cos of x plus two equals zero expand out i'm going to get six cos squared x minus three minus cos of x plus two equals zero and i'm just going to rearrange and simplify so 6 cos squared x and then minus cos x minus 3 plus 2 is minus 1 equals zero now going to factorize um and because we have a quadratic i'm gonna have three cos of x two cos of x and this is going to have to be minus one and plus one this equals zero so now for each bracket to equal zero i'm going to have cos of x equals negative a third and cos of x equals positive a half perfect so um remember i completely forgot to say this because i was overwhelmed by the quadratic but look at the interval within degrees calculator needs to be in degrees please so cos of minus the third and cos of x equals a half so i'm going to use the cast diagram so first up okay so we know this is 0 180 90. so cos of x equals a half we know that um the inverse is 60 degrees so cos to the minus 1 is 60 so we know this is 60 we're going to reflect it here so we know this is 360 minus 60 which is also going to give us 300 perfect and then we're going to have cos of x equals minus a third minus a third when i put that into my calculator um i'm going to get 109 okay which is roughly here so it's this angle and then i'm going to reflect it in this line if i do 360 minus 109.5 i get 250.5 now what i always like to do at the end of these types of questions is i like to write them out in ascendant order so x can equal 60 109.5 250 and 300 degrees so here are a few more examples and again i suggest you just try these on your own before you watch um let me go through them so we know that free a can be written as 2a plus a okay and we're going to use that fact to help us with this question so show that sine of three a equals three sine a minus four sine cubed a okay so first things first sine of 3a is going to be rewritten as sine of 2a plus a so using the formula of sine of a plus b we're going to get sine of a which is 2a now i know it says 2a and a b you've got to remember this is our a this is our b so sine of a and when i refer to a in this case i'm referring to this one and when i refer to being referring to this one so when i read the formula out i have sine of a oh black cos of b which in our case is a plus cos of a which in our case is 2a sine of b which in our case is a so this is just substituted into the formula now we also know that it wants us to show that we have at the end um four sine cubed and sine so we have two double angles here we need to make sure for cos because we have three different identities that we're just using the one that contains a sign so then sine of 2a we know this is a double angle formula and this is a double angle formula so we're going to also use sine of 2 a equals 2 sine a cos a we're going to use cos of 2 a equals 1 minus 2 sine squared a so instead of writing sine of 2a we're going to write 2 sine a cos a and then cos a plus and instead of writing cos of 2a we're going to write 1 minus 2 sine squared a and then we're going to have sine a now i'm going to expand and simplify and so 2 sine a and then cos a cos a is in fact cos squared a now we actually know that cos squared a equals 1 minus sine squared a so i'm going to substitute that in in a second and then i'm going to get plus sine a minus 2 sine cubed a and that's by expanding 1 times sine a and minus 2 sine squared times sine a so again i'm now going to substitute in um for cos squared a we're going to substitute in so this equals 2 sine a and cos squared we know is 1 minus sine squared a and then we still have plus sine a minus 2 sine cubed a so if i expand simplify i'm going to get 2 sine a minus 2 sine cubed a plus sine a minus 2 sine cubed a okay so if i collect all my sine a's i've got 2 sine a plus sine a which is 3 sine a got minus 2 sine cubed a minus 2 sine cubed a is minus 4 sine cubed a i think that's what it asked for yep okay so therefore proved okay and i've also kept a note here of the identities i've used and i think whilst we are using these types of questions we um should um write these down it'll help us as well with spotting patterns and what to use okay now part b says hence or otherwise if it says hence or otherwise what we need to do is use what we've just found out so if i have a look at what was in question one question one said sine of three a equals three sine a minus four sine cubed a okay oh and again look at our interval we're in radians now make sure you calculate a straight in radians and then they've told us 16 sine cubed theta minus 12 sine of theta minus 2 3 equals zero okay so i'm going to look at what they gave us in question one and what they've done in question two now straight away i can see we have a sine cubed sine cubed sine theta sine theta okay the sines are different but 16 and four it's been multiplied by four three and twelve been multiplied by four but this is now a negative so it's been multiplied by minus four so what i'm going to do is for this second equation here i'm going to take out a factor of minus four if i take out a factor of minus four minus four multiplied by what is 16 sine squared theta is in fact minus four sine sorry cubed not squared and then minus 4 times what is minus 12 sine theta that is in fact plus 3 sine theta minus 2 root 3 equals 0. i'm just going to rearrange the inside of this bracket so you can see i'm going to have minus 4 and then 3 sine theta minus 4 sine cubed theta minus 2 root 3 equals 0. now we know that freestyle theta minus four sine cubed theta here can be written as sine of three theta i know i can't believe it so minus 4 bracket sine of 3 theta close bracket minus 2 root 3 equals 0. okay this is really lovely now actually because now i can add 2 root 3 to both sides so i'm going to get minus 4 sine root 3 and we're back here to pure 2 equals 2 root 3 to get sine of 3 theta on its own i'm going to divide by minus 4 so we're going to get minus root 3 over 2. perfect okay so now we're ready to solve this our interval was between 0 theta is less than two pi so our interval was between zero is less than theta is less than two pi did it equal that no okay and then now if we're if our interval is between three theta and multiply everything by three we're gonna have six pi so sine um sorry so that means that three theta equals inverse sine of minus three root two which is in fact four pi over three as you know we're going to use the cast diagram to actually see um how many times if we think this is 0 2 pi pi so 4 pi over 3 is po 2 so 4 pi over 3 is going to be out here isn't it because it's one and a third pie okay to this whole angle it's 4 pi over 3 3 over 3 pi is going to be here so this must mean that this little angle here is in fact pi over 3. so now um obviously c a s t so now we know that's pi over three and we're going to have to take away if we reflect it in the vertical line here this is going to have to be pi over 3. okay so now we start at zero okay so we start at zero we go around the first time we're going to intersect is 4 pi over 3 because if i add um pi over 3 to pi i get 4 pi over 3. perfect then i continue to go round and the second time i hit it is here which is 2 pi minus pi over 3 which is in fact 5 pi over 3. then i continue to go round now remember now when we go around we're starting here this is now we're starting at 2 pi so we get here so this should be 3 pi so i've got 3 pi plus pi over 3 which is 10 pi over 3 and then i continue to go round stop here remember now we're getting close to 4 pi so 4 pi minus pi over 3 11 pi over three and then we continue to go round because our interval stops at six pi all the way around so that's four pi this now becomes five pi so five pi plus a third sixteen pi over three and then we carry on going round now remember this is now six pi so we're gonna do six pi minus a third which is 17 pi over 3 and then we stop because we reach 6 pi so now to find out what theta is i need to divide all of these by three which is going to mean they're all going to be over nine so here we have all of our solutions oh there we go okay nice and simple question right then um next one here we have solve four cos a theta minus 30 degrees equals eight root two sine theta in the range of zero is less than equal to theta is less than 360. degrees step one what step one degrees put your calculator into degrees please okay step two so here we have the addition formula so remember this is our a and this is our b so we're gonna have four and we're gonna have cos a cos b so cos theta cos b which is 30 and then it's a minus so it's plus four i mean you could always if you wanted to put the brackets then expand it but um and then this is going to be sine a sine b so that's sine theta sine 30 and this equals 8 root 2 sine theta so what am i actually going to do now well i know first of all that cos 30 is actually root 3 over 2. so everywhere there's a cos 30. i'm going to write this here okay and i know that sine 30 is in fact a half okay so if you know the exact values always substitute them in so we're going to get 4 cos theta times root 3 of 2 plus 4 sine theta times a half equals 8 root 2 sine theta so now what i'm going to do is i'm going to multiply and simplify each one of these normally i'd do this in my head and i wouldn't write the step down but i think it's good for you to see how we've done that so root 3 over 2 times 4 is going to be 2 root 3 cos theta plus 2 sine theta equals 8 root 2 sine theta so now what i'm going to do is um put my signs on one side have my cousins on the other and then i have sign and cos of course it's going to be tan so 2 root 3 cos theta equals 8 root 2 sine theta minus 2 sine theta if i factorize out a sine theta i'm going to get 2 root 3 cos theta equals 8 root 2 minus 2 lots of sine theta i'm now going to divide both sides by cos 2 root 3 equals 8 root 2 minus 2 sine theta over cos theta which we know are going to cancel to give me a ton and again i'm writing more steps than i normally would because i'd normally straight away just divide by this number and divide by cos so that we're going to have um it would just save time but tan theta i want you to see all the steps and then 2 root 3 over 8 root 2 minus 2 equals tan theta so now i'm going to put that into my calculator and tan theta equals 20.4 degrees again we have this in the interval so i'm going to draw my cast diagram okay 20.4 degrees is here i'm going to reflect it here so this is 20.4 we know this is zero and this is 180 so we're gonna have to do 180 plus 20.4 because if i start at zero go all the way around and that is the only other time it will intersect between zero and 360. so we're gonna have 180 plus 20.4 which is 200.4 degrees okay perfect so it's always good just some little tips to write down what you're substituting in okay write down your exact trig values okay what i would like you to do is stop and pause this and test your understanding and see how you finish these questions here are the answers please check through them um and make sure that if you um watch out for the plus and minus when we're square rooting here okay now you need to move on to exercise 4d and you need to do question 2 question 8 question 11 question 12 and question 14. so here's a sketch of y equals three sine x plus four cos x what do you notice about this sketch have a think about this please so you're right it has been translated and it's been stretched so if we think about a function if we think about a function f of x if it's been translated we know we're going to have f of x plus or minus um a number i'm just going to call this a and then if it's been stretched we know then that it's been multiplied the whole function has been multiplied so i'm going to write b on the outside now using trig we can actually we have another um identity here we can actually represent this equation in the form y equals r sine of x plus alpha where alpha is um you know how much we translate it by okay so if we have a a graph in degrees we would put degrees here if it was in radians we put radians here normally we you know we'd write it in terms of pi as well so a typical question would be put 3 sine x plus 4 cos x in the form r sine x plus alpha given alpha in degrees to one decimal place now we're going to be using the additional addition formula for sine of x plus alpha now remember this is our a this is our b so the addition formula now you can do r of all of that and then multiply each one by r but i'm just going to multiply it all by r so we know then that r sine of x plus alpha equals r sine a which is in fact x cos b which is in fact alpha plus r oh sorry r cos a which is x sine b which is alpha okay so i've already done that then i have a look and what i'm going to do is something called comparing coefficients so if i have a look here i have 3 sine x so that means if i compare that to 3 sine x and i compare that to where we have sine x we have r sine x cos alpha well if i compare that they both have the sine x perfect and then this one has three and this one has r cos theta so that must mean that free equals r cos um sorry not theta alpha okay and then i can do the exact same if i have a look at 4 cos x and i compare it with this one because we have a cos x we have r cos x sine alpha so again if i compare this they both have a cos x and then we have four and psi alpha so that must mean 4 equals sine r sine alpha so we know that r squared sine theta alpha plus r squared cos theta alpha equals r squared and how do we know that because sine squared x plus cos squared x equals 1. so if i multiply everything by r squared i'm going to have r squared sine squared x plus r squared cos squared x equals r squared okay remember if you're multiplying everything by and we multiply by r squared because r squared is a positive number okay and if we multiply everything by a positive number here we're not going to change the function and we're not going to change the function at all okay so all we're doing is stretching it so then um using the fact that r squared sine squared alpha plus r squared cos squared alpha equals r squared so we know that three equals r cos alpha from here so if i square this i'm going to get so we're going to have r sine squared alpha no i'm going to have r sine alpha squared plus r cos alpha squared equals r squared we know that r sine alpha is in fact equal to four so we're going to have 4 squared plus our cos output cos alpha equals 3. 3 squared equals r squared so that means r is in fact the square root of 4 squared plus 3 squared which is five okay now for the next step so step four using the fact that r cos sine theta over r cos theta equals tan theta okay so we know that r cos theta is in fact three and our sine sorry not theta alpha alpha so that must mean that tan of alpha equals three quarters so we can alpha is going to then equal 53.1 degrees to one decimal place so now i have all my information that i need i have r which is 5 i have alpha which is 53.1 i can now write it in this form so there we go r which is 5 sine of x plus 53.1 degrees we have now written this is what 3 sine x plus 4 cos x is equivalent to now why do we do this so we do this to help us solve because we've written it as a single trigonometric function so this is going to help us to solve so sometimes if we have 3 sine x plus 4 cos x no matter how much manipulation it's going to be quite difficult for us to then solve it i mean you could you know you know that cos of x is the square root of 1 minus sine squared x and you could substitute that in here but you know it just gets a bit messy this is a really nice way to then to be able to solve this type of equation it's always good to check isn't it so give it a go follow the steps and pause it here and test your understanding okay so here we go then step by step so step one we want it in the form r sine of x plus alpha in terms of pi okay so we know that our sine x plus alpha using the addition formula is going to be r sine x cos alpha remember because this is your a and b yeah and then it's plus r cos x sine alpha okay and i've multiplied both terms by r because it's multiplied by r here okay now what we're going to do is compare coefficients so here i have sine of x and here i have r sine x cos alpha so if i compare this this is one multiplied by sine x this is r cos alpha multiplied by sine x so i'm just going to write this down so this can be rewritten as 1 multiplied by sine x this can be written as r cos alpha multiplied by sine x so therefore if we compare the coefficients we both have a sine x and then this is one and this is r cos alpha so that must mean that one equals r cos alpha okay so one equals r cos alpha and then similarly we can apply the same principle to the other side here we have one times cos of x and here we have r sine alpha times cos of x so if we compare coefficients we have cos of x cos of x and the coefficient of cos of x is one so that must mean that r sine alpha must equal one okay so then we have one equals r sine alpha okay so that means that tan alpha equals silica which is r sine alpha over r cos alpha which in this case is one over one so tan alpha equals one so how do we find alpha tan minus one of one is going to give us alpha so our alpha equals make sure your calculator's in radians please okay and then we get pi over four perfect so we know what alpha now equals okay we'll manipulate it okay so we have our alpha x is our unknown anyway we have sine we now need to find r okay does anyone remember how we find r or why we find r okay so i'm going to use a different color so using the fact that we know sine squared plus cos squared equals one okay that should mean so using this principle we know that we could then write this squared plus this squared okay should in fact equal r squared because what's the difference between cos alpha and this first of all we've multiplied it by r okay and if i multiply everything by r here i now have r r okay so i multiply everything by r first okay then i square everything so we're going to have r sine squared plus r cos squared equals r squared so that's going to lead me to r squared sine squared alpha plus r squared cos squared alpha equals r squared okay now we already know what r cos alpha is and that's 1. so we're saying 1 squared plus 1 squared equals r squared so 1 squared plus 1 squared equals r squared so r equals the square root of 1 squared plus 1 which is 2. so now i have all the information i need i have alpha equals pi over four and i have r equals root two so for the last step i fill in so r which is root 2 sine lots of x plus our alpha which is pi over 4 and all i've done is rewritten this expression in this form for example you know when you have a quadratic and we complete the square yeah the quadratic is still the same when we complete the square we're just rewriting it as something else and the reason why we rewrite it in this form is that so it's easier for us to solve with the knowledge that we have in pure free so that's what we're doing we're manipulating this equation and changing it into this form they still mean the exact same thing but now we are able to solve this okay right so for the second question we have pet sine x minus root three cos x in the form r sine x minus alpha so our objective here is to find r and to find alpha and again it's in terms of pi okay so on page 71 step one we're going to rewrite r sine of x minus alpha and we're going to write that as sine x cos alpha minus cos x sine alpha okay we should all be happy with that and then we've got oh sorry and we're going to write an r because we've multiplied everything by r so we've got in our equation sine x multiplied by one and then we have um cos x multiplied by root three okay we look at we compare coefficients so they both have a sine x so that means in this case r cos alpha equals one so that's our first thing so r cos alpha equals one and then on the other side of the equation they both have cos x so that means r sine alpha equals root three because that's the coefficient here so we're gonna have r sine alpha equals root three so now how what can i do to find alpha well i know if i divide them by each other i'm going to get tan of alpha equals r sine alpha over r cos alpha so we know that the two r's are going to cancel and then we obviously know that sine alpha over cos alpha is tan alpha okay so again i'm going to have root three over one so to find so tan alpha then equals root three so alpha equals ten to the minus one of root three and we're going to use radians so in this case then alpha equals pi over three so check so we found our alpha now we need to find r okay and using the fact from earlier so in the exam just so you know to find alpha i would um always state my coefficients so i would write this part and then i'd write tan alpha equals this okay and then obviously we would have what alpha equals okay perfect now i would now i need to find my r and again it's using the fact that it's going to be um i know that r sine alpha squared plus r cos alpha squared equals r squared okay so i'm literally just going to do the square root of this squared plus this squared okay so i'm going to have the square root so i know that r equals the square root of 1 squared plus root 3 squared okay now 1 squared is 1 root 3 squared is 3. so this is actually root 4 okay when square root 4 is in fact 2. so now i have what r equals my last thing is to write it now in this form 2 sine x minus our alpha which is pi over 3. and there we go finished okay so it's just about learning this method i mean understanding why you know i've spoke to you about how we're manipulating it and how we use the knowledge of what tan is and that sine squared plus cos squared equals one to manipulate this but it's remembering this method so that you know and understanding why we do it so that then you can use it okay so here are the answers did you make sure you follow the steps again these are the kind of minimum steps you need to write down um please let me know if you need any more help yeah and we got it right right and now we have further examples which i'm going to go through so pet 2 cos theta plus 5 sine theta in the form r cos theta minus alpha where 0 is less than alpha is less than 90 degrees hence solve the equation where this equals free and again this is what i was speaking about earlier once we put it into the form r cos theta minus alpha we're going to be able to solve this equation a lot easier so here we go so we know that um by comparing coefficients we know that cos r cos theta minus alpha using the double angle formula is going to equal r cos theta cos alpha plus r sine theta sine alpha now equating coefficients we have cos theta so r so if i compare 2 cos theta by r cos theta cos alpha i'm going to have 2 equals r cos alpha and then for sine i'm going to get 5 equals r sine alpha so now um using r sine squared no r squared cos squared plus r squared sine squared equals r squared i'm going to have the square root r is going to equal the square root of 2 squared plus 5 squared so r equals root 29. then we're going to have um tan alpha equals sine over cos which is 5 over 2 so that means alpha equals 68.2 degrees all right so now i'm going to rewrite this so we're going to have r oh sorry we know what r is it's root 29 so root 29 cos theta minus 68.2 degrees okay now we have that we know this equals three so i'm going to rearrange so i'm going to have now that cos of theta minus 68.2 degrees equals 3 over root 29. when i have a look at three over route 29 um you have to remember now this is slightly different to when we have the interval when we're multiplying so now we have zero is less than theta is less than 360. now we have theta minus 68.2 is less than so on the left if we minus 68.2 we've got minus 68.2 and then 360 minus 68.2 is 290. um one point eight eight point two no wait there um two hundred 290 2.8 nope 291.8 perfect so this is our new range of values for when we're using the cast diagram okay so um when i inverse i'm going to get theta minus 68.2 equals minus 56.1 degrees okay so this is 56.1 here we have 56.1 right so if i really think about what's going on here now this is a zero and the first time i cross is 56.1 okay so i'm gonna write that down then i go all the way around and i cross here remember this is 360 minus 56.1 which is going to give us you know 300 and 3.9 the 303.9 is not in our interval so we don't need that one we're going to reject it but we do go back to minus 68.2 now lucky for us it gave us the negative anyway but normally then i'd go at zero and i'd go backwards and this would give us a minus 56.1 so um i only have two solutions for this and then now i need to add 68.2 to both sides so i'm going to get 12.1 degrees and 124.3 degrees and that is our final answer so now i really want you to give this one a try so find the maximum value of 12 cos theta plus 5 sine theta and give the smallest positive value of theta at which this arises so a bit of a clue um please make sure that you know we have cos and sine so you can use either r sine alpha plus theta or you could use r cos uh sorry theta minus alpha okay so you can decide whichever one you want then you have to think about what is going to give you the maximum or the maximum so you know for example that if you were to choose cos or sign but if we choose cos we know that cos of x has a maximum at one okay and the same with sine okay they both have a maximum one so if you had used r cos of these minus alpha cos of x has a maximum of one because it's one lot of cos of x so it would now gonna have a maximum it would now have a maximum at r and same if you did sine if you just used sine of x this would now have a maximum at one so if you did r sine of x it would now have a maximum at r as well because we've multiplied the function by r so if you know the maximum is that r you also need to think about well when does the maximum occur because we know the maximum occurs when x equals zero and for sine we know the maximum occurs when x equals 90 um if we're working in degrees okay so in either case once you've put it into this formula you know the maximum occurs when x inside the bracket for cos equals zero so you'd put theta minus alpha equals zero and then you'd find what feature equals which you would find in this case would be the maximum would occur at plus alpha so if we're looking at cos the maximum occurs at alpha one then if we're looking at sine it's going to be when theta plus alpha equals one uh no sorry 90. so we know that our x coordinate is going to be 90 minus alpha and our maximum is 1 which is sorry which is r okay so it wants the smallest possible theta value um these are just the coordinates but the smallest possible theta value for the maximum is either going to be alpha or 90 minus alpha depending on which maximum um oh 13 should be the maximums of both is it 90 minus alpha right well try both ways and see what you get so um the answer is 22.6 so give that a go let me know how you get on right then so we're going to have some quick fire of the maximums so we know the maximum of sine theta is 20 and the smallest possible value is 90 degrees because if you think about the curve this always occurs at 90 and then for cos 270 0 360. remember we want the smallest possible value so really we're just looking so 5 minus 10 sine theta okay so inside of outside the bracket outside we've multiplied one we've multiplied everything by -10 okay and then we've added five so our minimum is going to it's going to reflect so now think about the graph now if i multiply this is now going to be we're going to reflect it because i've multiplied it this is then going to be 10 but then i've added 5 so this is now 15 and this is going to be at 270. so the maximum is at 15 and the smallest possible value the first time this occurs is at 217 degrees because here our maximum is going to be three okay and this is theta plus 20. so we're going to take away 20 so that means this is going to move here so our smallest positive is actually going to be 360 plus um minus 20 which is 340 degrees okay so we've got a bit of a challenge for the last one do you think you can do it have a go um and then let me know okay so i really want you to further test your understanding okay and have a look at what we could do here so first of all um express six cos theta plus eight sine theta in the form r cos theta minus alpha okay so expressing that form and make sure that your alpha is three decimal places then for part b you're going to calculate the maximum value now here you have six cos theta plus eight sine theta which was used in question one so please rewrite that as r cos theta minus alpha from there you'll be able to find your maximum okay and the value of theta at which the maximum occurs so give that a please answers are here let me know what you got so please can you now do question seven question nine question 10 question 12 and question 15. okay and then the last thing is we're going to be proving trigonometric identities so then prove that 2 tan theta equals 2 over cot theta minus tan theta so um what we're going to do is make sure that we use the appropriate double angle formula or addition formula so tan of 2 theta so left hand side equals tan of two theta now using the formula we know this equals two tan theta over one minus tan squared theta okay not really sure where where to go from here but what i need to do is look at the question so it's two okay and if i look at here how have they gone from 210 feet to two they've divided by tan theta okay well let's see what's one divided by tangent theta ah that does give me cot theta and tan feet tan squared theta divided by tan theta will give me tampering with that okay nice and easy by comparing my answer to what i've done i now know they've divided everything by tan theta so because we have to prove i need to show that step so i'm going to write divide tan theta and then i'm going to show it 210 theta over tan theta all over 1 over tan theta minus tan squared theta over tan theta so now i'm going to have the tan fetus cancel which is going to give me 2 over 1 over tan theta cot theta okay and i might even just write that um and then tan squared theta over tan theta is just going to give me sorry that's a minus minus tan theta which equals right hand side therefore proved okay and because it's an identity you can put three lines okay next one so we're going to prove that 1 minus cos 2 theta over sine two theta is equivalent to tan of theta okay so i'm not sure which double angle to formula to use yet because of two theta because that's three okay um so i'm just going to quickly write down sine of two theta there's only one double angle formula so i have two sine theta cos theta oh before i do that sorry right left hand side equals 1 minus cos of 2 theta over sine of two theta which equals sine of two theta i know the double angle formula is two sine theta cos theta okay now remember i'm trying to get to ten now with the cos double angle formula so we have cos 2 theta equals there's three of them there's 1 minus 2 cos squared theta 1 minus 2 sine squared theta and i can't remember the other one it has a cos squared theta and a sine squared theta in it oh it's um cos squared theta minus sine squared theta okay so if i think about it how do i get tan silica so i want to get rid of the sine and i want a sine over cos okay so cos two theta if i want to get rid of this sign i'm gonna have to have sine squared and i want to sign and a cos okay yeah i'm gonna go for 1 minus 2 sine squared and hope for the best so 1 minus bracket 1 minus 2 sine squared theta okay which is going to equal 1 minus 1 plus 2 sine squared theta over 2 sine theta cos theta okay 1 minus 1 cancels so i'm then going to have oh and 2 divided by 2 is going to cancel so i'm now going to have sine squared theta over sine theta cos theta sine squared theta i can rewrite just to show you but i wouldn't do this in the exam sine theta sine theta which we know is then going to cancel which is going to give you side theta over cos theta which is tan theta therefore proved now the more the more proven questions we do the more you're going to know which one to substitute in and which one's going to work um for the manipulation that we want so i want you to test your understanding and give these a go now these are from the ocr exam board so i know we're not with ocr but they are some good questions now um just give it a go and see how you get on and remember what you're trying to achieve so remember what cot x is and if you've got you know you're trying to show that cot 2 of x plus cos 2 of x is completely equivalent to x okay then for the second one by writing cos of x equals cos of two times x over two so this is actually a double angle formula because we're doing two lots of something so this is cos two a okay and your a is x over two in this case okay so they've given you a clue um please try your best okay so here are the answers how did you get on these are the absolute minimum steps required for these preview questions so please do not miss out steps pause it now so you can see the answers now what i'd like you to do is go to um on exercise for if you could complete question four six eight nine and ten so question six we've got show that question so we need to show that cos four theta is three over eight plus a half cos two theta plus one over eight cos four theta okay step one now again with these questions you never know what's gonna happen okay so i'm gonna look in the if you go to page in the book if you go to page 96 we have a summary of all the formulas okay so page 96 we've got the addition we've got double angle and we've got the r sign or our cos okay so keep on page 96 while you look at this now the first thing really i see i see that i've got a double angle formula okay um and i've got a double angle formula twice really well three times right this is what i'm going to do i have no idea if this is correct but i'm going to roll with it okay and intuition i want cos to the powerful so i'm going to use the double angle formula here and i'm going to use the one that just has a cos squared in okay and the same i'm going to use a double angle formula here now i always write down what formula i use on the side i just think it's good practice so i'm going to be using cos of 2 a equals 2 cos squared a minus 1 because i i want to get cos to the power 4. so i'm you know logically why would i substitute in the one that says cos and sine because i know that even though cos equals the square root of one minus sine squared it's just going to get messy okay so number one intuition number two let's really think about this so cos of two theta we know is going to then become 2 cos squared alpha minus 1. so ins so i'm going to have 3 over 8 as normal plus a half bracket and then instead of cos of 2 theta i'm going to write 2 cos squared theta minus 1 okay straight substitution okay um and then i'm going to have plus 1 over and i'm going to substitute again but imagine so i've got four theta okay so this means that my a is now two theta okay so i'm to put though so then i'm going to have um 2 cos squared 2 theta minus 1. right i'm going to expand out everything first okay so i'm going to have 3 over a and then a half times 2 cos squared theta is in fact cos squared theta oh sorry and i'm going to have a half times negative 1 is minus a half okay everyone should be happy with this first exception then the second expansion 1 over 8 times 2 cos squared 2 theta so 2 cos squared to theta is going to give me um it's a quarter yeah plus a quarter cos squared two theta minus an eighth okay so here i am right i'm just going to sort this out a little bit three over eight three over eight minus an eighth is two over eight which is a quarter minus a half i think i'm going to get negative a quarter plus cos squared theta plus a quarter cos squared two theta so this is where i'm at okay now what i'm going to do i'm going to look at these formulas again and see if there's anything i can do right cos squared to feta hmm so cos squared two theta okay actually can be written like this cos of two theta or squared okay again we have another double angle formula and i'm going to substitute this one back in again so let's see so now i'm going to have plus a quarter of 2 cos squared um theta minus one all squared and then obviously i have my minus a quarter cos squared theta okay great so now what do i have to do so now i've got to expand this bracket okay so now i'm going to have 2 cos squared theta minus 1 2 cos squared theta minus one and then obviously i'm going to multiply the whole thing by a quarter after okay so but let's just forget about that for now so when we expand the bracket two cos squared theta times two cos squared theta is four cos to the power four theta minus one times minus one is plus one minus one times two cos squared theta is minus 2 cos squared theta and then 2 cos squared theta minus times minus 1 again is minus 2 cos squared theta okay then now i'm gonna then this is gonna give me four cos four to the power of theta oh that's a plus um plus one minus four cos squared theta okay is everyone happy that's what it expands to now i'm going to multiply it by a quarter aren't we so now if i bring all this down i'm going to use blueprint so we have a minus a quarter plus cos squared theta plus and now i'm going to do a quarter times this whole bracket a quarter times four cos to the power of four theta is just cos to the power of four theta a quarter times one is a quarter and a quarter times minus four cos squared theta is just minus cos squared theta okay this is really good everyone because guess what now minus a quarter plus a quarter is going to cancel cos squared theta minus cos gross theta is going to cancel and what are we left with cos for theta which equals the left hand side therefore proved oh right hand side equals okay so remember this is a six mark exam question okay and the only rule that i've used is this double angle formula and i've just kept substituting it in until it makes sense now this is where your intuition really comes into play okay and sometimes you do go wrong and sometimes you think oh but is that right but again you've got to remember that cos squared of two theta can be rewritten as cos of two theta all squared okay so okay step one use your intuition okay and you think you've got cos to the power of four and then you've got two theta and a four theta so you've clearly got some double angle formulas yeah so i look at the double angle formula cos to the power four i can't do anything with that i don't really know any rules that are going to help me so i know i'm not going to start on that side i'm going to start on the right hand side and i've got a double angle formula but because i want a cos to the power four i'm going to find the double angle formula which just has cos in it okay so there's no purpose sign so it's like a hunch right now i have three over eight three over eight plus a half a half instead of writing cos 2 theta what am i going to do i'm going to substitute this in okay so i'm going to substitute in 2 cos squared and then here i need to half that to get my a because remember this is double it isn't it so half of 2 feet is just theta minus 1. so i've substituted that in here so instead of cos 2 theta it's 2 cos squared theta minus 1. okay then i've got plus 1 over 8. now here i've got cos 4 theta i know if i half that that means my a in this case is 2 theta okay so i've used the same formula but i've got my two feet so i've got so two cos squared two theta minus one okay now i expanded all my brackets so free over eight check a half times two cos squared theta is cos squared theta a half times minus one minus a half one over eight times two cos squared two theta a quarter cos square two theta one over eight times minus one minus one over eight okay perfect i'm now here i'm now looking at this thinking how on earth am i going to get anywhere no idea yet so all i did in the next step was simplify i had three over eight and minus a half and minus an eighth so i just simplified all my numerical values okay which gave me minus quarter and then i had a cos squared theta here and then i had a quarter cos square two theta here now i'm looking at the formulas i've got and none of these got squared in and then i remember okay i remember that cos squared 2 theta is actually can be written as cos of 2 theta all squared that's what it means isn't it so if you if you knew that cos of 2 ft equals 4 well cos squared of 2 theta would be 4 squared okay so with this knowing this i now know that cos squared two theta okay can be written as cos of two theta squared cos of two theta is two cos squared theta minus one so i substituted that in here okay and then i squared it okay then i've done two cos squared theta times two cos squared theta four cos to the power of four theta minus one times minus one plus one minus one times two cos squared theta minus two cos squared theta two cos squared theta minus times minus one minus two cos square root of theta i then simplified this minus two cos squared theta minus two cos squared theta minus four cos squared theta okay so i now have four cos four theta plus one minus four cos square theta that is what this equals and now i'm going to multiply it by a quarter okay so a quarter times four cos four theta is cos four theta a quarter times one is a quarter a quarter times minus four cos square theta is minus cos squared theta and then i brought this down so i'm gonna if i rewrite this in so i'm gonna have here minus a quarter plus cos squared theta plus a quarter of this bracket okay that's what happened i brought that down okay and then there's like a little step in here this is this little step in here is here okay then lastly i simplified it minus a quarter plus a quarter is zero cos squared theta minus cos squares theta is zero and then i'm left with cos to the power four theta which equals my left hand side of the equation therefore proved and then finish off with a final challenge for you please pause it now i'm not going to give the answer to this challenge and let's see who completes it