in this video i'm going to go over the precalculus second semester final review so in our first unit we talked about trig identities and part of that was verifying where you're trying to show one side equals the other so the first task is to decide which side so looking at this there's more things i can do here so that's going to be the side i'm going to choose then to turn everything in terms of sine and cosine so i'm going to change this tangent into sine squared theta over cosine squared theta [Music] and that one's just sine still okay next i want to get a common denominator so i'm going to multiply this one by cosine squared over cosine squared [Music] so this will be sine squared theta cosine squared theta all over cosine squared [Music] okay now that we have a common denominator we can combine these things [Music] okay next i can factor out a sine squared so i'm going to have sine squared theta which will leave me with one here minus cosine squared over there [Music] one minus cosine squared is sine squared [Music] okay then i want to look at what i'm trying to show i'm trying to show tangent times sine so what i'm going to do is just pair these two together and make this sine squared theta over cosine squared theta times sine squared which is tangent squared [Music] okay and this one uh everything's in terms of sine and cosine this we can do more things with so i need to get a common denominator so i'm going to multiply this one by one minus sign over one minus sign and multiply this one by one plus sign over one plus sign okay which gives me this then i'm going to combine them and i'm going to distribute through the cosines sorry i realized i forgot the original denominators there anyway so we distribute through the cosines and then when we get this denominator when we foil it you're going to have 1 plus sine of x minus sine of x minus sine squared so when we combine everything the this and this cancel this and this cancel so in the top i have 2 cosine x in the denominator we have 1 minus sine squared which of course is cosine squared so this reduces to 2 over cosine of x which is equal to 2 secant of x so getting a common denominator and then just working your way through it all right next we have the binomial theorem so the binomial theorem the first step is to take whatever row of pascal's triangle we are interested in so you will be given a pascal's triangle uh so if this is four we want the fourth row which is going to be one [Music] four six four and one so i do one of those kind of as a placeholder next we take whatever is first so the x squared raise it to this power and then repeat that in descending order so three two one and none then we'll take the second thing which is negative 2y to increasing powers so there's not one here negative 2y here negative 2y squared negative 2y cubed and negative two y to the fourth now we need to simplify so we'll do this in a couple steps here uh first um we have the power to the power so that's gonna be x to the 8th here this negative is being raised to an odd power so this next term is going to be negative and then i'll multiply the 2 and the 4 to get 8 and then we have x to the 6th and y this next one i'm going to do in two steps so this is going to be 6 x to the fourth and then negative 2 y squared is going to be 4 y squared and it's positive the next one is going to be r odd power so that makes this negative 4 x squared times 2 i cubed is going to be negative i already dealt with that 8 y cubed and then this will be a positive 16 y to the fourth so my last answer is x to the eighth minus eight x to the sixth y six and four make twenty-four 4 and 8 is 32 [Music] and 16 y to the fourth so your final answer should uh alternate because of the negative we've got the powers of two don't forget to raise those powers and then so each term should have a coefficient x to some power and y to some power [Music] in vectors so what we have here a plane uh travels at 275 miles per hour at a bearing of 310 there is a wind blowing out of the so blowing at a bearing of 65 degrees at 30 and this is o65 not six so it's um so first i'm going to draw vector diagrams of each [Music] okay so we have a bearing of three tens so bearings due north is zero that's 90 180 to 70. so 310 is going to be over here okay so this is the 310. what we're interested in is the reference angle so this is going to be 40 because it's going to be 40 degrees past there for the wind it's at a bearing of 65 so 65 is going to be like here this is the 65 the reference angle that we need is 25. next we're going to write the component form [Music] okay so you can use whatever notation just as long as it's clear what's happening in the x direction and what's happening in the y so this one is going to be a negative positive so my first is going to be negative we do the magnitude which is 275 times the cosine of the reference angle which is 40. my y component and i'm going to use i the i and j notation [Music] okay the vertical piece is positive and then it's going to be 275 sine of 40. and i'm going to label that with a j again you can label it as long as it is clear what is x and what is y now we're going to type it into the calculator and make sure your calculator is in degrees okay so that's what's happening with the plane it goes negative 210 in the x direction positive 175 in the y this next one the reference angle we said was 25 and this is going to be a positive positive because it is in fact in the first quadrant okay so we have the magnitude cosine 25 magnitude sine 25 again using the i and j type it into the calculator so that is our wind vector now we are going to add so you're going to add this and this we're going to add these two numbers and these two numbers so you add the x's add the y's so this is the component of the sun now we're going to turn it back into what's actually happening so what is the true ground speed which is going to be the magnitude of this vector and what the bearing is okay so for this we're going to do the square root of this squared plus this squared okay so this is our ground speed [Music] for the bearing we're going to start with tangent inverse so i'm going to do the tangent inverse of the y over the x or the 189.446 [Music] over and i'm going to ignore the negative 183.473 this gives us an acute angle of 45.918 now we have a little work to do here [Music] my result was a negative positive so that means we're going negative 1 8 3 here and 1 8 9 here so this is my resultant vector the 45.9 that we just found is the reference angle is that we want the bearing from due north so in this case we're going to take that number and add it to 270. so my true bearing is 315 degrees 0.918 [Music] last free response problem we have a power series which is an infinite geometric series the common ratio we're multiplying to get from term to term here is one third x so we know we can find the sum of an infinite series if the common ratio is between negative one and one so i'm going to take negative 1 put 1 3 x in the middle here and then just multiply by 3. so my interval of convergence is negative 3 to 3. then the rational function is just a 1 over 1 minus r so a one is three over one minus one third x and you can leave it just like that okay next we have another verifying problem so these are the multiple choice questions uh these are all no calculators so keep that in mind as you go uh so the first thing i'm going to do is get a common denominator so multiply this by sine plus y over sine plus y multiply this one by sine minus y over sine minus y okay i don't have a whole lot of room here so we're going to work sideways so sine minus sine is zero and then we're gonna have one minus negative one so on the top i have two when we foil this denominator we get sine squared minus one well this is equivalent to negative cosine squared [Music] okay which is equivalent to negative 2 secant squared for the next problem and the next problem after that we are going to be using the trigonometric identities so on your formula sheet for this one we're going to be using the sum and difference identities specifically the difference for sine okay so the sum and difference identities sine a plus or minus is this whole thing uh so since this is minus this will be minus so i'm going to write that out first [Music] so there's the general formula uh keep in mind these are all multiple choice so the amount of work you show is up to you then i'm going to substitute in pi for a and theta for b [Music] so we get that then you're going to want to use your unit circle to evaluate those to the pi's so sine of pi the y value here is zero [Music] the y value [Music] sorry the x value there is negative 1. so out of all of this the negative negative makes this a positive so this is equal to sine of theta in this next section we're going to be using the double angle identities so on your formula sheet we're using the double angle identities we're going to be using sine cosine and i'm actually not going to use the one for tangent and we'll see why [Music] first sine 2 theta is 2 sine theta cosine theta well i know what cosine of theta is i do not know what sine of theta is but if you recall we figured out that we can find this using a triangle so i'm going to call that angle theta and then here if cosine of theta is 5 over 13 i'm going to say the adjacent side is 5 and my hypotenuse is 13 and i want to find that okay so you might know this pythagorean triple but if you don't we can use the pythagorean theorem so 25 plus x squared equals 169 subtract to get 144 so x is 12. [Music] so my new piece of information is that sine of theta is going to be opposite over hypotenuse so 12 over 13. so now we just put those in here sine is 12 over 13 cosine is 5 over 13. [Music] so 2 times 5 is 10 so we get 120 over 169 okay next from the formula sheet cosine 2 theta there's three options i'm just going to pick the first one which is cosine squared theta minus sine squared theta then i will substitute my cosine [Music] squared minus my 12 over 13 squared so i'm going to leave some space for tangent so i'm going to work sideways so we're going to have 25 over 169 minus 144 over 169 [Music] which is a negative 119 over 169 okay last for tangent [Music] there is a formula but once we know sine and cosine this is just going to equal the sine of two theta over the cosine of two theta okay which is going to be 120 over negative 119. [Music] next we have solving problems which relies on a lot of the same things we've been using um first looking at this i've got both sine and cosine which is not ideal so my first step is going to be taking this cosine and turning it into a 1 minus sine squared [Music] then i'll distribute through so we have 2 minus 2 sine squared x minus sine of x plus 1. well the 2 and the 1 are going to be 3 but i don't like the fact that i've got a negative in front of these so i'm going to move it all over here so that'll be a positive 2 sine squared a positive sign and a negative 3. okay well now easy to figure out what factors will give us this so it's going to be a 2 sine theta and a sine theta one of these has to be uh positive and negative if they multiply to be three and they have to be three and one uh so running through the different possibilities if i have a plus three minus one that middle term is negative two plus three which gets the positive 1. so either 2 sine theta plus 3 equals 0 or sine theta minus 1 equals 0 which means sine of theta equals negative 3 over 2 or sine of theta equals 1. well between 0 and 360 sine is never less than 1. so that one does not have a solution sine is 1 only once at the top of the table at 90 degrees [Music] the last one quite simple cosine of x is one half for zero to 360 you're looking at your unit circle uh the x values are one-half at 60 degrees and at 300 degrees [Music] okay next we're going to draw some vector diagrams and find the reference angle so standard position [Music] is our unit circle measurements so the positive x-axis is 0 90 180 270. so this one standard positions 140 is going to be over here [Music] this measurement is 140. the reference angle is the angle from the x-axis which is going to be this angle here which is going to be 40. so our reference angle is 40 degrees [Music] bearing and we've already seen this with the plane problem is measured from due north so this is zero this is 90 this is 180 this is a bearing of 270. so a bearing of 300 goes all the way around here this is 300 my reference angle is this angle which is going to be 30. [Music] [Music] okay so next we're going to figure out uh we're going to make numbers here so that the vectors are perpendicular and so they are parallel well if they're perpendicular that means their dot product equals zero the dot product is the product of the x's so negative one times k plus four times 12 and we want that to equal zero [Music] so my k value will be 48 to make perpendicular vectors [Music] for parallel what i'm going to do is switch one of these and change the sign of 1. so i'm going to make this negative 12 k that would be perpendicular these two are perpendicular so if these two are perpendicular and these two are perpendicular these two are parallel so now i'll do the dot product with those two so negative 1 times negative 12 plus [Music] uh plus 4 times k set that equal to zero so twelve plus four k equals zero four k equals negative twelve which means k equals negative three okay next we're going to identify the ordered pair that goes from this vector point to this point and then we're going to find the magnitude well just think about taking a trip if i start at negative 9 and i go to 1 that means my x coordinate is going to be 10 because we're going positive 10 from negative 9 to 1. if i'm going from negative 1 to negative 3 that means i'm going down 2 so my y component component's negative 2. the magnitude is just the square root of 10 squared plus 4 squared sorry 2 squared which is four so this is going to be square root of 104. okay so these next two i'm going to kind of talk about together we want to write the parametric equation of one and slope intercept and then we're going to write the slope intercept form of something in parametric so this is what a parametric equation looks like the general formula is x equals x1 because i'm just going to call it delta we added at a1 t y is y1 plus a2 t uh a point on the line is any x1 y1 so any points on the line can go there and then a1 a2 is our direction vector so this is going to be my change in x this is my change in y so putting that with the line i know my slope [Music] is i'm going to put the negative with the 4. it actually doesn't matter which well that is delta x over delta y [Music] okay so for my lines we're going to have an x equals and a y equals we're going to pick any point on the line well there's an infinite number but the easiest is going to be the y-intercept and then i'm going to add my change in apologies this is float rise over run right guys delta y over delta x so my change in x is five my change in y is negative four this would be one option keep in mind this is multiple choice the negative could have gone with the five which would make this plus that would be valid but these are one example you just have to find the right one on multiple choice here we're going to kind of work backwards okay so i know my point on this line is six 7. the slope is going to be delta y over delta x let's not make that mistake again so my change in y is 5 my change in x is 8. then what i'm going to do is actually put this in the point-slope form of a line which is y minus 7 equals 5 8 x minus 6. so y minus 7 equals 5 8 x minus 30 eighths so 15 fourths and then i'm going to add seven which would be twenty eight fourths [Music] so 28 minus 15 would give me a 13 fourths okay next we're going to graph in polar uh so i'm going to call this point a to 120. so you're going to go out to and then go over to 120. [Music] when we have a negative positive you can go to 7 pi over 4 which would be here and then go negative 3 or you can go to negative three and go clock counterclockwise seven pi over three either way we're going to end up here call this one c when we have a positive negative rather than going 120 this direction we're gonna go 120 this direction [Music] and go up one three [Music] and then the last one a negative negative you can go to negative pi over six and then go negative two okay or you can go to negative two and go clockwise that measurement okay next we're going to be graphing uh lines and circles so just basic equations negative 310 is going to be the same as positive 50 degrees this would be 45 so negative 310 is going to be just a little bit more than that [Music] okay so negative 310 would be this measurement here but it's all the same line r equals three you're just going to go out to 1 2 3 and graph that line or circle next we're going to be converting between rectangular coordinates and polar and we're going to look both directions so going from rectangular to polar the r value is going to be the square root of the x squared plus y squared theta is going to be the tangent inverse of y over x in the correct quadrant [Music] so let's do some of these so first our r is going to be the square root of root 3 over 2 squared plus 1 half squared that's going to be the square root of three fourths plus one fourth which is four fourths which is one square root of one is of course one theta is going to be the tangent inverse of one half over root three over two which when you do this math you end up with tangent inverse of root three over three we want this in quadrant one so tangent is root three over three at pi over six okay not a bad idea to put those tangent values on your page so my coordinate is one [Music] pi over six the second one it's a little tough to tell but these are supposed to be negative negative so this is quadrant three okay so we'll do our r value is the square root of root two squared plus root two squared which will be two plus 2 which is 4 square root of 4 is of course 2. theta is going to be the tangent inverse of negative root 2 over negative root two which is tangent inverse of positive one which occurs at both pi over four and in the third quadrant five pi over four so my coordinates here are 2 5 pi over 4. when we're going the other direction x equals r cosine theta and y equals our sine of theta those are our two conversion equations so super easy x is equal to 3 cosine 45 y is equal to 3 sine 45. use your unit circle cosine of 45 is root two over two the sine is also root two over two so multiply those together i'll get my coordinates three root two over two three root two over two [Music] okay this one x equals negative 2 cosine negative pi over 3 y is going to equal negative 2 sine negative pi over three okay negative pi over three is just down clockwise pi over three so it's going to be equivalent to 5 pi over 3. so the x value there is one half [Music] so this is negative 1. the y value there is positive sorry negative root three over two so this is going to be a positive root 3. so we get the coordinates negative 1 and positive root 3. okay in this next section we're going to be converting equations rather than coordinates if we have a polar equation we want to try to get our sine theta or cosine theta which i can get by multiplying sine just like with the coordinates our cosine r sine theta is one so this is y equals five this one i'm going to first start by foiling this part [Music] okay the nine and nine cancel so i'm going to write this as x squared plus y squared equals negative 6 y then i'm going to replace this with r squared and this with r sine theta and then i can divide both sides by r again keep in mind this is multiple choice [Music] but that's our answer next we have the polar form of complex numbers so when we're multiplying dividing or doing powers when you're multiplying you multiply the two r values so i'm going to multiply 3 and 4 and we're going to add the two theta values so just to get a common denominator this is two pi over six so i'm going to add seven pi over six plus two pi over six [Music] again this is multiple choice so the amount of work you showed is really up to you so 3 times 4 is 12 and then we have cosine nine pi over six which on a multiple choice is not going to be this labeled it'll be the simplified so divide both by three to get three pi over two [Music] okay for division you're going to divide the r's and divide the thetas okay so i'm going to take c1 over c2 so we're going to have 3 over 4 and then we're going to have cosine [Music] theta seven pi over six minus the other theta [Music] okay so this ends up being three-fourths cosine five pi over six which happens to be one of our unit circle values [Music] okay we want to answer in rectangular form so i need to go ahead and simplify this uh my x value at 5 pi over 6 is going to be negative root three over two my y value is going to be positive one half and then we'll go ahead and distribute so this is gonna be a negative 3 root 3 over 8 plus 3 over 8 i the next is going to be the reverse of that but we want our answer in polar form so we have 4 over 3 and then when we do 2 pi over 6 minus 7 pi over 6 we get 5 pi over 6. then we want to write the equivalent positive form so our final answer here would be 4 3 cosine 7 pi over 6 plus i sine 7 pi over 6. so you can maybe think about this graphically or just add 2 pi which would be 12 pi over 6. okay next we're going to use powers so i'm going to take c1 and raise it to the fourth so we'll take the r value which is 3 raise that to the fourth power then we're gonna multiply the theta which is seven pi over six times four so we get 28 pi over six [Music] okay so we've done the work here but now we need to get this into rectangular form three to the fourth is eighty-one first i'm going to reduce this fraction and make this 14 [Music] pi over 3 [Music] okay which is not on the unit circle okay so what i'm going to do is subtract 2 pi which in this case is going to equal 6 pi over 3. okay so in this case i'm going to do that twice i'm going to 6 pi over 3 would get me to 8 pi do that again to get 2 pi over 3. [Music] okay [Music] so then we use our unit circle cosine of 2 pi over 3 is negative one-half the y-value there is positive root 3 over 2. so my final answer here is going to be negative 81 halves plus 81 root 3 over 2. [Music] our next unit was sequences in series and we started off with factorials so when we're simplifying you want to take the bigger thing and write it in terms of the smaller one so the numerator is just going to be n minus 1 factorial for factorials we have the first number times the number one less than that which would be n times the number one less than that which would be n minus one factorial [Music] so those will cancel and i'm left with 1 over this times this which would be n squared plus n would be your multiple choice answer you're looking for [Music] here we can rewrite this as n plus 1 times n factorial over n plus one so this simplifies to just n factorial [Music] okay we're going to look at different situations and decide if it is a permutation or a combination in a combination the order does not matter permutation order does matter so save has a four digit combination to unlock it if numbers cannot be repeated how many combinations are there this is ironically order does matter you can't put them in just a random order they must be in the correct order so a combination lock is actually a permutation okay an omelette station is running especially on omelettes how many three item omelets can made from 10 ingredients well it doesn't matter what order we put them in this is a combination [Music] in the olympic lew's competition there are 32 10 23 competitors how many ways can gold silver and bronze be awarded order definitely matters here and then the lottery uh it doesn't matter what order the numbers are picked just that you've picked the correct six numbers so this is a combination [Music] okay next we're going to write an x recursive and explicit uh so recursive we always want to say what a 1 is and then we say how we want to get to the next term so here we're doing is we're subtracting 5. so to get to a n plus 1 we're going to take the term before it and subtract 5. explicit you want to be able to plug in 1 2 3 4 to get those values but if we know this a specific type so this one is arithmetic we can just use those nth term formulas so my a n would be a1 and these are on your formula sheet plus n minus 1 times my common difference which when you simplify this because again this is going to be multiple choice is going to be negative 5 n plus 9. [Music] the next one is geometric because we are multiplying so still for the recursive a1 is negative one a n plus one we are taking the previous term and multiplying it by negative three so negative three times a n for the explicit the general formula is a n equals a 1 which is negative 1 times the common ratio which is negative 3 to the n and that's good for simplifying you could write it as negative this but the negatives don't multiply so be careful there okay so we're going to find a three and a six for two different types of equations one is recursive one is explicit so for the recursive to get a3 you need a2 and so on so we actually have to do the whole thing here uh a2 is going to be 3 minus 2 times a1 which is negative 1. so that will be 3 plus 2 which is 5. a3 is going to be 3 minus 2 of that 5 so 3 minus 10 is negative 7 a4 is going to be 3 minus two times negative seven so i'll be fourteen plus three or seventeen a five will be three minus two [Music] seventeen and a6 is 3 minus 2 times negative 31 which is 65. so our two answers are [Music] uh but if it's explicit all we have to do is plug in the number so a3 here is going to be three times three minus one so three times two is six to get a six we have six times six minus one so six times five which is thirty [Music] okay next we're looking at are they arithmetic geometric or neither so this one is just kind of bouncing around it's neither this one we're multiplying by two to two so since we're multiplying by two this is geometric here we're adding two so this is arithmetic [Music] uh so in an arithmetic series if a 3 equals 7 and a 7 equals 37 what is a1 [Music] which is seven a four we don't know a5 a6 and a7 is 37. so what i'm gonna do for this is re number [Music] make this a one which makes that a5 okay so in my nth term formula i'm going to say 37 equals a1 plus 5 minus 1 times my common difference [Music] so i can subtract 30 equals 4d so my common difference is 30 thirty fourths or fifteen halves okay then i'm gonna go back to my original numbering structure uh and either plug in this one or this one it doesn't matter so if i say seven equals a one plus if i'm plugging in seven that means that's the third term so n is three [Music] sorry common difference is fifteen halves [Music] so it's going to be 2 times 15 halves which is 15. so my a1 is negative eight next we want to form a geometric sequence with one geometry mean between two so a1 is 2 a2 is some number i don't know and then a3 will be 72 so we're going to use the 72 equals a1 r to the n minus 1. so 36 equals r squared so r is 6 so then if i multiply this by 6 this would be 12 and that would be my geometric sequence okay next we get into sigma notation so here i don't have a function versus this one i do so when i plug in 2 we have 5 when we plug in 3 we get 5 when we plug in 4 we get 5 5 we get 5 6 we get 5. seven we get five okay so we have six fives or a sum of thirty expand this one so this is two times six plus two times seven plus two times eight is two times nine plus two times ten it could be written like this or twelve plus 14 plus 16 plus 18 plus 20. um notice it doesn't say to find the sum it just says to expand so this we expanded this next one right in summation notation so there's different ways we can do this [Music] okay so i'm going to start with my general formula we've got 2 on the top here and then we have the odd numbers so the equation 2 and either plus 1 or minus 1 will give me the odd numbers if i do minus 1 that means i'm starting at 1 because 2 times 1 minus 1 would be 1 and you can always double check and this is plus dot dot dot so two infinity okay here we want to find the sum of the infinite series if it exists which means we need to look at the common ratio here we're multiplying by three over four so my ratio is three over four which means the sum certainly does exist and is a one over one minus r so it's going to be two-thirds over one minus three-fourths so that's two-thirds over one-fourth which is going to be eight thirds when we multiply by the reciprocal this one my common ratio we're multiplying by four on the top two on the bottom and it's alternating so negative four over two so there is no sum because that is definitely not between negative 1 and 1. in this section we're determining whether the type of series and whether it is convergent or divergent so here two-thirds one-third one-sixth is infinite geometric okay the ratio is one-half which means this is convergent [Music] okay here this would be two over one two over two two over three two over four which is the harmonic series which is always divergent [Music] okay this next one we're multiplying by 4 on the top and 2 on the bottom which is an infinite geometric however this one is divergent [Music] this one we're multiplying by two on the top and two on the bottom so our ratio is one uh so this is actually divergent and it would be geometric or arithmetic i guess [Music] okay next we have limits so for these polynomials we're looking at the degree so the degree is the same so it's going to be the ratio of the leading coefficient here the degree is bigger on the top so we would say this one does not exist this one the degree is bigger on the bottom so it goes to zero for these exponentials if the base is between negative one and one it goes to zero if it's bigger than one then it goes to infinity or does not exist e is about two point seven one eight pi is about three point one four so that number is less than one so that is zero so that is your pre-calculus final review you