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Verifying Trigonometric Identities



in this article we're going to talk about how to verify trigonometric identities which is not a very easy topic but here's what you need to do typically you'll be given a problem with an equation and what you need to do is you need to show that the left side of this equation is equivalent to the right side of this equation now listed on the board are some techniques that you can use to do just that one of the first things that i would look to is trying to convert everything to sine and cosine that's a good place to start it doesn't work for every problem but at least for some of the easier ones it's a a good way to get started solving it now for step two sometimes you'll see a problem where there's two terms on the left but one term on the right when you see this what you could try to do is if a and b are fractions you can try to get the common denominator of both fractions convert it into a single fraction and then go from there other times you may need to factor so that you can convert the two terms into one term now step three is basically the reverse of step two if you have one term on the left and two terms on the right to convert from one to two terms you could distribute or if you have a fraction you can split that one fraction into two smaller fractions and for step four converting division into multiplication or vice versa for that I have to show it to you you just have to see it in action number five sometimes you need to multiply by the conjugate if you see one plus sign it could be one minus sign or one plus or minus cosine let's say on the denominator of a fraction that's a good indication that you need to multiply by the conjugate and then for step six sometimes you need to factor other times you need to foil and if you've factored in algebra then you can apply those same techniques here sometimes you'll see situations where you have difference differences of perfect squares or you may have a trinomial like this where you can see that it's factorable so those are some other things that you want to keep in mind when solving these problems now before we begin working on a few practice problems you may want to take down some notes by the way I recommend writing what you see on the board right now so the first thing you need to know is some identities sine squared plus cosine squared is equal to one make sure you know that one that's a very common one and then i'm sure you've seen this one 1 plus tangent squared is equal to secant squared so you want to commit these identities to memory because you're going to be using them a lot the next one 1 plus cotangent squared is equal to cosecant squared now you need to be familiar with these other identities tangent is equal to sine over cosine cotangent is the reverse it's cosine divided by sine so therefore tangent is the reciprocal of cotangent tangent is one over cotangent and then here are some other reciprocal identities secant is equal to one over cosine and cosecant is one over sine so make sure you write this write all this stuff down on a piece of paper it's going to be very helpful as we work on some problems let's begin with the first problem let's say we have sine x times secant x and let's say that's equal to tangent x go ahead and verify this trigonometric identity by the way for each of these problems i recommend that you pause the video try it yourself and then you know after you finish working a problem play the video to see if you have the right answer so we're going to use step number one we're going to convert everything into sine and cosine now you could start with the left side of the equation and make it look like you like the right side of the equation or you could start with the right side and then make that equivalent to the left side or you could start with both and make make them equal to each other it really doesn't matter as long as you show that the left side is equal to the right side you'll get the right answer but for me personally i like to keep the right side the same and I like to start with the left side of the equation and convert it to the right side so you don't have to do it that way but that's the way I like to do it which that's the way i'm going to be using in this video so just to give you a heads up so right now i'm going to leave the sine function the way it is now we know that secant is one over cosine now if you want to you can write this as sine x over one sine times one is just sine and on the bottom if we multiply these two one times cosine is cosine and based on the formulas that we wrote earlier we know that sine divided by cosine is tangent so now that the left side is the same as the right side we're finished here so we've verified this particular trigonometric identity so once you show that the left side is equal to the right side you're finished now let's move on to our next example number two let's say we have tangent squared times cotangent squared and that's equal to one go ahead and verify this trigonometric identity so let's begin by converting everything into sine and cosine which is step number one so we know that tangent is sine over cosine now we have tangent squared so we have two of them so this is going to be sine over cosine squared cotangent is the reciprocal of tangent we know that cotangent is cosine over sine and this is squared as well so what i'm going to do is i'm going to distribute the exponent this is sine to the first power and it's raised to the second power so 1 times 2 is 2 so this becomes sine squared and the same is true of cosine we're going to raise cosine to the second power so this becomes cosine squared times cosine squared divided by sine squared now cosine squared divided by cosine squared that's one sine squared divided by sine squared is one so basically we get one times one which is one so that's it for number two that's all that we need to do in order to verify this particular identity now let's move on to our next example number three so for this one here's the problem cotangent times secant x times sine x is equal to one go ahead and verify it so like before we're going to use step number one we're going to try converting everything into sine and cosine so we can change cotangent into cosine over sine secant we know is one over cosine and sine we could just leave it as sine x or you can write it as sine x over one so these two will cancel sine x over sine x is one and cosine x divided by cosine is one so this works out to be something similar one times one is one and so that's it for that problem all right let's work on some harder examples each problem will progressively get harder this one is a little bit harder than the last one but not too hard so here we have cosine times secant divided by cotangent and this is equal to tangent go ahead and prove it so let's begin by converting everything on the left side into sine and cosine so cosine i'm just going to leave it the way it is i'm going to write it as cosine over 1. secant we can write that as 1 over cosine now cotangent if you want to you could convert it to cosine over sine you don't have to but for this problem actually i'm going to leave it as cotangent and here's why we know we can cancel cosine and so that's going to give us a one in the numerator and we know that one over cotangent is tangent so for this problem there's no need to convert cotangent into cosine of a side if you did convert it you can still get the right answer so it wouldn't be a bad thing you'd just be introducing an extra step so the last thing to do is replace one of a cotan with tangent and that'll be it for this problem now let's move on to our fifth example for this one we're going to have sine x times tangent x and this is going to be equal to 1 minus cosine squared divided by cosine x so what do you think we need to do for this problem how can we change sine times tangent into 1 minus cosine squared over cosine so sometimes what you could do is look at the right side to see where to go the right side of the equation if you're starting from the left side that is the right side of the equation can be like a guide and we only have the cosine function on the right side so that tells us we need to introduce cosine into the left side right now the only thing we could change is tangent we can convert tangent into sine and cosine so i'm going to write sine as sine over one and tangent i'm going to replace that with sine over cosine now if you don't want to rewrite what's on the right side you could just put this now let's multiply sine x times sine x is sine squared and on the bottom we have one times cosine which is cosine so the good thing is we have what is on the bottom so this part is the same what we need to do is convert sine squared into one minus cosine squared now how can we do that well we're familiar with this identity we know that sine squared plus cosine squared is equal to one so what i'm going to do is i'm going to subtract both sides by cosine squared so these two will cancel and i'm going to get sine squared is one minus cosine squared so knowing that all I need to do is replace sine squared with an equivalent expression 1 minus cosine squared and that's going to give me the answer and so now the identity has been verified try this one cosine squared minus sine squared let's say that's equal to 1 minus 2 sine squared so what do we need to do here so once again we're going to look at the right side of the equation as a guide on the right side there is no cosine function only sine so somehow we need to convert cosine into a sine function and we know how to do that using this identity cosine squared plus sine squared is equal to one so we need to get cosine squared by itself so for this time we need to subtract both sides by sine squared so on the left side what we have left over is cosine squared and on the right side we have 1 minus sine squared so let's replace cosine squared with 1 minus sine squared so we're goin g to have 1 minus sine squared minus another sine squared and that's equal to what we see on the right side now there's like an invisible one here so we have negative one minus one which is negative two so then this becomes one minus two sine squared and that's all we need to do for this problem so it helps to look at the right side which tells you what you need to do with the left side so use the right side of the equation as a guide all right number seven so this one is going to be a little different than the others here we have sine x times tangent x plus cosine x and this one is equal to secant x so notice that we have two terms on the left one term on the right so this is like the first term sine times tangent and here this is the second term cosine and this is the third one so we have this form a plus b equals c what do we need to do here well we can't factor out a sine or cosine or tangent because we don't have a common factor so the other thing that we could do is somehow get two fractions get common denominators and then combine it into a single fraction the question is how can we do that well let's begin by turning everything into sine and cosine that is everything on the left side so let's write sine as sine of one and tangent we're going to convert that into sine over cosine and then cosine x we're just going to write it as cosine over one we want to put everything in fraction form on the left we can multiply sine times sine which will give us sine squared on the bottom one times cosine so that's just cosine so now we have two fractions but we don't have the same denominator how can we get common denominators in order to do that we need to multiply the second fraction by the denominator of the first and whatever you do to the top you must also do to the bottom so if you multiply the numerator of this fraction by cosine you need to do the same thing with the denominator so we're going to have sine squared over cosine x plus so cosine times cosine that's cosine squared and on the bottom one times cosine we know it's cosine all right let me delete a few things just to create extra space so now that these two fractions share the same denominator what we can do is combine it into a single fraction so we can now write sine squared plus cosine squared all divided by the same common denominator of cosine now hopefully you have your list of trigonometric identities with you because we have an important identity one that you're familiar with sine squared plus cosine squared what does that equal but we know that sine squared plus cosine squared is equal to one so now this becomes one over cosine and we know that secant is one over cosine so we can replace one over cosine with secant and that's it for this problem the identity has been verified so whenever you see a situation like this where you have two terms on the left one term on the right you may need to convert it into two fractions get common denominators combine it into a single fraction and then simplify now let's move on to number eight here we have secant minus cosine x and this is equal to tangent x times sine x go ahead and try it so notice that we have two terms on the left well this time it's a minus b instead of a plus b and we only have one term on the right tangent times sine that's one term so we need to do something similar either we could try to factor or we can try to convert into fractions and get common denominators we don't have a common term to factor so we need to get common denominators so let's turn secant and cosine into fractions secant we know it's 1 over cosine and cosine we can turn that into a fraction by putting it over one so now in order to get common denominators we'll need to multiply the second fraction by cosine over cosine so this becomes one over cosine minus cosine squared over cosine so now that we have the same denominator we can combine this into a single fraction so this becomes 1 minus cosine squared over cosine x and that's equal to tangent times sine x now what should we do at this point notice that this 1 minus cosine squared is an identity we know that sine squared plus cosine squared is equal to one so if we take this term move it to the other side we're going to get sine squared is equal to one minus cosine squared it's positive on the left side but when you move it to the other side it's going to be negative so we're going to do is we're going to replace 1 minus cosine squared with sine squared so we have sine squared over cosine and this is equal to tangent times sine now notice what we have here this is basically rule number four here we have division and here we have multiplication how do we convert division into multiplication well let me show you let's say if you have a over b you can convert that fraction into multiplication by writing it this way a over one times one over b so that's a simple way in which you can convert division into multiplication so we can write this as sine squared over one times one minus cosine now sine squared is sine times sine now what you could do is you could move this sign to this fraction there's nothing wrong with that for instance let's say if I have 4 times 4 over 1 times one over two if I were to move the four to this other fraction it would be four over one times four over two the value of the entire situation is still the same this is 16 divided by two that's eight this is still four times four which is 16 divided by two that's still eight so you could move the sign from one fraction to the other when you're multiplying it doesn't change the value of the fraction so what we now have is sine x over one times sine over cosine and that is equal to tangent times sine x sine x over one we can just leave it as sine and sine divided by cosine we know it's tangent sine times tangent is equal to tangent times sine you can reverse it three times five for instance is equal to five times three they both equal 15 so once you get to this part that's it the problem is finished 

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