in this video we will be talking about the trig identities and also solving trigonometric equations but that will be a later video so um these identities should be very familiar to you from trigonometry so i do kind of have an expectation that you should probably memorize these guys um even though you will have your notes for exams um because of quarantine life but you should at least know some of the basic ones okay so i'm not gonna go ahead and drive them that probably you have seen already in intrigue but uh these are very easy to do the reciprocal quotient and pythagorean uh double angle are very useful especially when we're trying to reduce a couple of problems and even when we do some equations it's a little easier to work with the double angle or maybe if you want to go back to the single single theta sum and difference cofunctions these cofunctions you have probably seen a lot uh especially when we're looking at sine and cosine and when we're shifting and reflecting and then half angle and half angle identities are also pretty useful but the ones that are i think that are that you're going to really uh look at especially in calc 2 and calc 3 are the reduction formula so when you're reducing a power to a single power these are super duper important so if there's something that you need to really know it's these guys all of these guys are important but these are kind of like the most general ones that you really want to make sure that you know have handy um some some way or another because they will help out with a lot of the calculus that we do um now notice that there are some two of them that we i'm neglecting which is a sum to product and product to some formulas they exist um i don't really care for them so i mean you can go ahead and take a look at them in the book but i'm not going to go ahead and talk about those these are the ones that i really want to look at so what we're going to do is that we're going to take a look at some of these identities and um and do a little bit of problems and we're going to use each of the identities or the identity that best matches the problem that we that we are given so so let's do a couple of examples so here let's find the exact value of cosine of pi over 12. now uh one thing that you could do is you can just put cosine pi over 12 in the calculator but what i want to make sure that you know how to do also is be able to give the exact answer of the cosine of pi over 12. so um if you think about our unit circle so here's our unit circle uh you can see that pi over 12 is not on there but we know that and we got pi over six pi over four and pi over three so how can we possibly get pi over 12 from these three different values because pi over 12 is a little bit less than pi over six well if i take pi over 4 and subtract pi over 6 i will get pi over 12. so i can rewrite this as the cosine of pi over 4 minus pi over 6. and if i do that i will end up getting pi over 12. so now that i did it as a difference of two different angles i can go back up and use uh the sum and difference value for cosine so cosine alpha minus beta equals cosine alpha cosine beta plus sine alpha sine beta so just be careful with that plus these guys are are the opposite so when it's plus is going to be minus when it's minus it's going to be plus okay so in this case in this case we're going to have cosine so let me write it up here so cosine alpha minus beta is going to be cosine alpha cosine beta plus sine alpha sine beta so i'm just going to go ahead and plug in these angles into the formula so you're going to get cosine of pi over four then you're going to get the cosine of pi over six then plus the sine of pi over four times the sine of pi over six right so now uh we know that the cosine of pi over four is square root of two over two we know the cosine of pi over six that's going to be square root of three over 2 plus the sine of pi over 4 that's going to be the square root of 2 over 2 and the sine of pi over 6 that's going to be one half so what you're going to get is square root of 6 divided by 4 plus square root of 2 divided by 4 which gives me square root of 6 plus square root of 2 divided by 4. all right so this would be the exact answer of of cosine of pi over 12. so this is what you have to do kind of break them up with known angles and then uh use one of the sum sum and difference identities okay now let's go ahead and do cosine of 15 degrees using half angle formula so you can see that this guy you could have easily done maybe 45 degrees and subtract the 30 degrees to get the 15 degrees but now we want to use the half angle formulas so the cosine of 15 degrees so let's go back up and look at the half angle formulas and the half angle identities right here will be a cosine of alpha over two is equal to the plus or minus square root of one plus cosine of alpha divided by two so let's go ahead and write that down so we have the cosine alpha over two is equal to plus or minus square root one plus cosine of alpha divided by two okay just double check make sure that i did this right oh you don't want to mess this up okay all right now remember what we really want is to reduce this cosine of 15 degrees into an angle that we know we divide by 2 and that will give me 15 degrees so the only number that i can divide by 2 to get 15 is 30. so we're going to let alpha be our 30 degrees so this alpha let me make it into a highlighter this alpha is just going to be this 30 degrees so then i'm going to get the plus or minus square root of 1 plus the cosine of 30 degrees divided by 2. now notice here we can actually drop one of the plus or minus this just depends on what quadrant you're in because uh 15 degrees is in quadrant one it'll make sense for us to just use a positive sign so i'm just going to use the positive the positive square root so uh the cosine of 30 degrees is just going to be square root of three over two so i'm going to have the square root of one plus square root of three over two divided by two and then i can go ahead and and do a little bit of reduction so on the top you'll have two over two so two over two plus square root of three over two divided by this gives me square root of 2 plus square root of 3 divided by 2 over 2 which gives me the square root of 2 plus square root of 3 divided by 4 which in turn gives me the square root of 2 plus square root of 3 and then if i take the square root of 4 on the bottom i'm going to get 2. so this would be as reduced as it will get okay so this will be the exact answer for the cosine of 15 degrees all right now let's look at this next example so uh we are given the sine phi is equal to 5 over h and phi is between 0 and pi over 2. so let's find sine 2 v cosine 2 phi and tangent to v so it looks here that we're going to have to use a double angle identity so one of the things that i like to do is that i let you locate this sine phi of 5 over 8 on your coordinate axis so i'm gonna draw my x and y plane and then we know that phi is from zero to pi over two well that's just from zero to ninety degrees so we're gonna draw a triangle over here and here is a 90 degree angle here is phi we know that sine is the opposite over the hypotenuse so you're going to have 5 and then 8 over here now we need to figure out our angle a or our side a so we're going to use pythagorean so we're going to have a squared plus b squared equals c squared so our a is going to be what we're looking for our b is going to be 5 squared and our c is going to be 8 squared okay i don't know what that is 64 minus 25 gives me 39. so a equals a very ugly number square root of 39 okay all right so now that we know what a is which is square root of 39 and we know that we're going to look at the positive side because in quadrant one everything is positive uh we're going to need we need to use the formulas for the double angles to figure out the rest of the guys so let's start off with sine of two feet sine two phi oh for those that don't know this this symbol means v or phi so let's go back up and look at sine 2 v so we're going to look at the first one so 2 sine theta cosine theta so let's go ahead and write that down so this is going to be 2 sine theta cosine theta or going to be sine phi cosine v so we got a 2 there the sine phi we already know what that is that's 5 8. cosine phi that's going to be the adjacent over the hypotenuse which is the square root of 39 over eight so this is going to give me 10 square root of 39 divided by 64. and i think you can reduce that a little bit 10 over 64. that gives me 5 over 32. so 5 square root of 39 over 32. so this guy will be the sine of 2 feet now let's look at the cosine of two phi so if we go back up the cosine of two phi you actually have three different options they're all going to give you the same thing i'm just gonna look work with the first one cosine squared minus sine squared so i'm going to use cosine squared phi minus sine squared phi so the cosine we figured it out in the previous one the square root of 39 over 8 squared and the sine is just a 5 8 squared so this is going to give me 39 over 64 minus 25 over 64. okay so we're going to take that 39 subtract the 25 gives me 14 over 64. okay which also reduces to 7 over 32. and now we're going to look at the tangent of 2 2 phi so if you look at the tangent the double angle for the tangent is going to be 2 tangent over 1 minus tangent squared so two tangent over one minus tangent squared two tangent all right and if we look at the tangent the tangent's gonna be just the opposite over the adjacent so it's going to be 2 times 5 over square root of 39 divided by 1 minus 5 square root of 39 squared so on the top you're going to get 10 over square root of 39 you're going to get i'm going to make the 1 into a 39 over 39 and this guy's just going to be 25 over 39. this is going to be let me write it down here 10 over square root of 39 divided by so i'm going to take 39 subtract the 25 that's going to give me that 14 over 39 so what would this give me divided so flip so i'm going to get 10 over square root of 39 times 39 over 14. oh gosh i don't even know what that is okay so 10 times 39 and i got and i'm going to multiply this so divided by 14 so i get 195 over 7 times square root of 39 okay so not a pretty number so check my math make sure that everything i'm doing this writing okay but this is basically it so you're basically just plugging and chugging making sure that you use geometry to figure out the rest of the triangle right and then after that just plugging it in all right so now let's use the reduction formula to reduce cosine to the fourth power x now this might be seem pretty useless but it's actually really really useful in calc two so you'll be doing this a lot more in calc two so we're gonna do it to the fourth power okay so uh one way that we can reduce cosine to the fourth power right so notice that we want this so that there's no powers of sine or cosine greater than one so uh if you look at the formulas we don't have anything that has to the fourth power and the reduction formulas but is there a way that i can reduce the uh square the 4 0 squared so one thing that i can do is i can make this into cosine squared x times cosine squared x because this can give me to the fourth power and now i can apply the reduction formula so the reduction formula says that cosine squared x is equal to 1 plus cosine 2x divided by 2. okay so let's go ahead and do that so the first guy is going to be a 1 plus cosine 2x divided by 2 times 1 plus cosine 2x divided by 2. so i'm gonna notice that i get a fourth down here i'm going to pull that out and i get 1 plus cosine 2x 1 plus cosine 2x i'm going to foil all that out so i'm going to have 1 plus 2 cosine 2x and the last one is going to be cosine squared 2x all right so it looks like i'm basically done but notice that when i foil this thing out i ended up getting another cosine squared x all right so here's the tricky part because we don't have a cosine squared 2x but notice that if i have cosine squared x right notice that what happens is that when i have 1x here you have two x's here so basically the argument just doubles so you can see that this x to this 2x just doubled so that means cosine squared 2x is just going to be equal to 1 plus cosine 4x divided by 2. so you can see that here this guy is just going to be doubling the argument so we're going to use that that that detail so i'm going to have 1 4 1 plus 2 cosine 2x and then the cosine squared 2x is just going to be what we just figured out which is going to be that 1 plus cosine 4x divided by 2. so this guy's going to give me one fourth one plus two cosine two x plus one half plus one half cosine 4x so all i just did there was i just broke up this fraction i divided this 2 into both of them into the 1 and into the cosine 4x um and it seems that the only thing that i can just do is just add the one and the one half which gives me three halves so we have three halves plus two cosine two x plus one half cosine four x and i have no more squares everything is a cosine to a power of 1 everything is all good so you might be wondering what the heck was the whole purpose of this and you're actually not going to know the purpose until calc 2 but typically what's happening is that in calc 2 uh integrating this is so don't freak out but integrating something that looks like this is really difficult okay in the calc 2 area you can't really do calculus with this function this function is really hard to integrate however it may seem a little bit weird but actually integrating this guy is actually a lot easier than integrating something like that i know this looks a lot more complicated but actually in calculus this is a lot easier to work with than something to a power of four okay so uh we'll continue with uh with the other identities in the next video