[Music] in this video we're going to look at the exact trig values for S cos and tan you need to know the values that these give for the following angles 0° 30° 45° 60° and also 90° for example without using a calculator you might be asked to work out s of 30 which is equal to 12 you could also be asked tan of 60 which is equal to the < TK of 3 so since you're not allowed a calculator to answer these questions it will appear in paper one let's take a look at how we can learn these values so I like to draw a table that goes like this along the left hand side I write s cos and tan and along the top I write those angles so 0 30 45 60 and 90 now we're going to start with sign to do sign I'm just going to count from 0 up to four so 0 1 2 3 and 4 then we square root each of these numbers so squ < TK 0qu < TK 1qu < tk2 < TK 3 and squ < TK 4 and then we divide all of these by two now these are the correct values a few of them though will simplify so if we take a closer look at < TK 0/ 2 the square root of 0 is just 0 so here we actually have 0 over 2 and 0 / 2 is actually just 0o so your calculator will give you the value 0 if you did s of 0° let's have a look at this next one so in the next one we have squ < TK 1 / 2 square < TK of 1 is just 1 so this is actually 1/ two or a half and this is the value your calculator will give the next two values < tk2 over2 and S < tk3 over two are correct we don't need to do anything with those values but the final one we do Square < TK 4 over two the square < TK of 4 is two so this is actually just 2 over two and 2 / 2 is one so your calculator will give you the value of one if you do s of 90 so that's all of the values for sign done now really nicely the values for cos are the same just in Reverse so the one we got for S of 0 which was 0o is actually the answer to COS of 90 s of 30 which was 1 12 is the same as COS of 60 s of 45 which is < tk2 over 2 is the same as COS of 45 s of 60 is the same as COS of 30 and S of 90 is the same as COS of 0 so now we've done two of the rows we just need to do the row for Tan to work out the values for Tan we actually start in a very similar way that we did for S so we count from 0 to four so 0 1 2 3 4 and then we square root all of these but when we did sign we then divided all of them by two but we're not going to do that this time we do need to divide them though and we divide them by the same numbers bit in Reverse so < TK 4 < TK 3qu < tk2 squ < TK 1 and < TK 0 and then we just need to work out what these give so if we look at the first one squ < TK of 0 over Square < TK of 4 the square < TK of 0 is just 0 and the square < TK of 4 is 2 so we have 0 over 2 which is zero so T of 0 is 0o then if we move on to the next one we have < TK of 1 over < TK of 3 the S < TK of 1 is just 1 so we have 1 over the < TK of 3 and the < TK of 3 is a number that we can't work out we need the calculator for that so we just leave it as the square < TK of three then if we move on to the next one we have squ < TK of two over < TK of two now when you divide any number by itself you always get one for example 10 / 10 is 1 a million divide by a million is 1 even pi/ Pi is still one so the square root of 2 / the < TK of 2 is just 1 then if we move on to this next one we've got the square Ro T of 3 which we said we need to leave as the square Ro T of three but on the bottom the square Ro T of one which is one when you divide any number by one you just get the same number back for example 6 / 1 is 6 11/ 1 is 11 Pi / 1 is pi so the < TK of 3 / 1 is the < TK of 3 and now we we move on to the final value so here we have the square root of four which is two and on the bottom the square root of 0 which is 0 so we have 2 over 0 now you can't actually do two IDE by 0 it's not possible so this one is actually undefined and you can try it in your calculator it will give you a math error so you don't actually need to learn this value now I just want to return to the value of tan of 30 this one here 1 / < TK 3 is correct however sometimes this could be written as the < TK of 3 over 3 if you're studying the higher course then we've just rationalized the denominator on it it could be written as 1/ < tk3 or < tk3 over 3 so you might want to be aware of both ways of writing tan of 30 now let's have a look at how they assess this in exam questions so first of all and this is more common in a foundation paper they literally just ask you what the value of one of these is so they might say write down the value of s of 30 so if we look at our table we can see the S of 30 is 1/2 so we just write down 1/2 or they might say write down the value of COS of 90 so looking at the table that one Z or they could say write down the value of tan of 60 and looking at the table that one is < tk3 another type of question that you may need to do is finding missing values in a right angle triangle like this one here if you haven't already checked out my video on trigonometry using Sak TOA you may want to watch that video before doing this question so for this question here we want to find the value of x it's a right angle triangle so we can use soak TOA so we're going to label the sides the longest one hypothenuse is this one here the one opposite the angle in the question here is the opposite and the other one is the adjacent in this question we have the hypotenuse we're looking for the adjacent and we don't have any information about the opposite so we don't need to use the opposite so we need to use the adjacent and the hypotenuse so if we look at Sak TOA the one with the adjacent and the hypotenuse is this one here so we'd write out COS of theta is the adjacent over the hypotenuse now we substitu in the values so Theta is the angle in the question that's 60° the adjacent is the one we're finding that's X and the hypotenuse is 10 so we end up with cos 60 = x/ 10 now normally at this point we would multiply both sides by 10 and then reach for our calculator but this question would come up in paper one and that's because we've been given COS of 60 which is one of the values we need to know COS of 60 is equal to 1 12 so we can replace COS of 60 with2 now we just have an equation to solve so we could solve this by multiplying both sides by 10 or we could notice that to get from 2 to 10 you multiply by five so to get from 1 to X we also multiply by 5 and 1 * 5 is 5 so X will be 5 cm now let's try another example like this one so here we have this triangle and you'll notice that the angle is missing this time it's a right angle triangle so we can use Sak TOA so let's label the triangle the 22 is the hypotenuse the 11 is opposite the angle so that's the opposite and the other side is the adjacent we have the hypotenuse and the opposite but we don't have the adjacent so we're not going to need that in this question so we're looking for the one that has the opposite and the hypotenuse and that is this one here so we're going to use sign so we write down s of theta is the opposite over the hypotenuse and now we'll substitute in the values Theta is the angle which this time is X the opposite is 11 cm and the hypotenuse is 22 so we end up with sin of x = 11 / 22 you might notice that that fraction will simplify to 12 so we're looking for sin of x to be equal to 1/2 now normally we'd reach for our calculator and do the inverse sign of 1/2 but this will come up in paper 1 because you need to know which value of x for S will give You2 and we can find that by looking at the table so looking at the table we get 1/2 here which is when we do s of 30 this means that X must be 30 since s of 30 is a half so we can just write down that X must be 30° now if you study Foundation maths I wouldn't necessarily recommend watching the rest of this video I just go and try the exam questions that I put in this video's description but if you are a higher student you might want to take a closer look at where these values come from and you can derive them from triangles so we're going to take a look at how we can find all of those values from the table by looking at triangles we're going to start by looking at this right angle triangle here where the base and the height are both one if the base and the height are both one we can find the value of the hypotenuse using Pythagoras so if we label these sides as a and b and then the hypotenuse as C we know that a s + b s must be c^ 2 so 1 2 + 1 2 must be c^ 2 and if we do 1^ 2 and 1 s they're both 1 so 1 + 1 is c^ 2 and 1 + 1 is 2 so 2 = c^2 we can reverse the order of this so c^2 = 2 and then square root both sides so c will be the sare < TK of two so we now know the length of the hypotenuse it's the square < TK of two now since we have the base and the height being equal in length here this is actually an isoceles triangle so we can find the value of these two angles here since this angle is a right angle at 90° you can do 180 take away 90 which is also 90 and then divide that by 2 which gives you 45 so both of these angles are 45° now we only actually need one of these 45° angles so I'm going to remove the one at the top Now using this as our main angle we're going to do some Sak TOA so let's label the triangle this is the hypotenuse the one opposite the angle is the opposite and the other one is the adjacent so if we now write down the trig ratios for S cos and tan and substitute these values in the Theta is 45 in all of these the opposite is 1 the adjacent is also one and the hypotenuse was the < TK of 2 so we find that s of 45 is 1 / < tk2 so cos 45 and T of 45 is 1 over 1 we can do some simplifying for each of these so for the sin 45 and the COS 45 we're going to rationalize the denominator by multiplying the top and bottom by < tk2 on the top this will give us 1 * < tk2 which is < tk2 and on the bottom < tk2 * < tk2 which is 2 you obviously end up with the same value for cos so < tk2 / 2 then for T 45 we've got 1 over 1 and 1/ 1 is 1 so using this triangle here we can derive the values for S cos and tan of 45 and if I bring back the table from before you can see that they do match now let's have a look at another triangle so this time we're going to take this triangle here this this is an equilateral triangle this time so all of the sides are the same length and we're going to set them at two since it's equilateral all of the angles are the same so they must all be 60° now we're going to split the triangle in half with this line here this will create a right angle at the bottom and we'll split that angle at the top into two so each side will now be 30° also the base of the original triangle was two so the base of each of these two triangles is one we're only going to need to use one half of this equilateral triangle so we'll use this half here now we have one bit of information missing from this triangle we need to know its height and we can do that using Pythagoras again so if we label the triangle the two shorter sides is a and b and the hypotenuse is C we can do a S + b s is c^ 2 so a 2 + 1 S = 2^ 2 well 1 s is just 1 and 22 is 4 if we subtract one from both sides on the left hand side we'll end up with a^ squ and on the right hand side 4 takeway 1 is three then if we square root both sides we'll end up with a equaling the square < TK of three so let's label that height as the square < TK of three now we can use this triangle to work out even more values we're going to look at the 30° first so we're just going to remove the 60° temporarily so if we label this triangle the two will be the hypotenuse opposite the angle is the one so that's the opposite and the remaining side the squ < tk3 is the adjacent now let's write out the trig ratios for S cos and tan if we substitute in these values the is 30° the opposite is 1 the adjacent is < TK 3 and the hypotenuse is 2 so we end up with sin 30 is a half cos 30 is > 3 over2 and tan 30 is 1 over < tk3 for Tan 30 we can rationalize the denominator so on the top we get 1 * < tk3 which is < tk3 and on the bottom < tk3 * < tk3 which is 3 so tan3 is < tk3 over 3 now if we compare this to the values in the table you can see that these are correct now let's bring back the triangle from before and look at 60° instead now we need to switch around the opposite and the adjacent this time because opposite the 60° currently says the adjacent but that needs to be the opposite then if we write out the trig ratios for S cos and tan and substitute these values in so the angle Theta is 60° the opposite is < three the adjacent is one and the hypotenuse is two so we end up with these we only need to simplify the bottom One S < tk3 over 1 is the < TK of 3 and if we bring about that table you can see that these values are correct now what about the values for 0 and 90 well you can't really do Soaker TOA with a 0° angle or a 90° angle actually instead we can consider what happens to an angle As It Gets very very close to zero or very very close to 90 so let's label this angle here as Theta then if we write down s of theta is the opposite over the hypotenuse and we'll label on those two sides this one here is the opposite and this one here is the hypotenuse now watch what happens if we change this triangle so that Theta becomes a little bit bigger and bigger still and bigger still and even bigger still eventually what happens is that angle will get closer and closer and closer to 90° we can't actually get there otherwise the triangle would just turn into a straight line now let's consider what happens to the opposite and the hypotenuse as that angle gets closer and closer to 90 so if we just reverse this process and go back and then look closely at the opposite and the hypotenuse we know the hypotenuse is longer than the opposite because it's the longest side in the Triangle but as we continue to increase the size of the angle the hypotenuse gets closer and closer to the sides of the opposite as you can see here they're really close now if we could continue so that this triangle turned into a straight line then the opposite and the hypotenuse would have the same value so when we came to do the opposite divide by the hypotenuse we just be dividing a number by itself and dividing any number by itself it's one so we can use this to understand that the sign of 90 gets closer and closer to one now let's have a look at the same idea for cos so cos Theta is the adjacent divid by the hypotenuse so this time we're going to label those two sides the adjacent and the hypotenuse once again we'll increase this angle so it gets bigger and bigger and bigger and close to 90 so watch what happens to the two sides this time the adjacent is getting smaller and smaller and as we increase Theta so it gets so big eventually the adjacent will disappear all together and become zero so COS of 90 must be 0 over whatever the hypotenuse is and 0/ by any number is zero now let's have a look at what happens in terms of tan so for Tan Theta we need the opposite and the adjacent so let's label the opposite and the adjacent now we'll increase the size of theta so it gets closer and closer to 90 and you can see that this time the opposite stays the same length but the adjacent gets smaller and smaller and eventually the adjacent would become zero so we'd have the opposite whatever that length is divide by zero and we said earlier that you can't divide a number by zero which is precisely why the value of tan of 90 is undefined a number that we can't work out if we take a look at the table you can see that these numbers match so s of 90 is 1 COS of 90 is 0 and tan of 90 is undefined now let's have a look at how we can work out the values for 0° so if we take this triangle again and we're going to start with s so we'll need the opposite and the hypotenuse but this time we're going to make that angle smaller and smaller until it gets close to zero so let's start to shrink the size of the angle down and pay attention to what happens to the opposite and hypotenuse you can see here that both of them are getting smaller and smaller but the hypotenuse will eventually become the same length as the adjacent but the opposite will go to zero so in terms of the opposite divided by the hypotenuse will have 0o over some number and 0 divid by any number is zero so the sign of 0 is zero now let's have a look at cause so for cause we need the adjacent and the hypotenuse so let's label those sides and we'll start to make that angle smaller and smaller and smaller so this time the adjacent doesn't change size at all but the hypotenuse does it shrinks down and eventually if we made that angle as small as possible the adjacent and the hypotenuse would be the same length and when you divide two numbers that are the same by each other you end up with one so this is demonstrating that COS of zero will become one and finally we need to look at tan of zero so for Tan we need the opposite and the adjacent so let's label those and we're going to shrink that angle down so this time the adjacent doesn't change size but the opposite gets smaller and smaller and eventually if we got smaller enough that opposite would go to zero so we end up with 0 over the adjacent and 0id by any number is 0o so tan of 0 just like s is also zero if we now bring back the table you can see these numbers also match s of 0 is 0 COS of 0 is 1 and tan of 0 is 0 thank you for watching this video I hope you found it useful check out the one I think you should watch next subscribe so you don't miss out on future videos and why not go and try the exam questions in this video's description