Proving Trigonometry Identities
Date: 01/07/97 at 01:27:26 From: Baine Subject: analytic trigonometry I just started a new chapter on analytic trigonometry and am really stuck. The problem is to prove the following identity: (tan x+cot x)tan x = sec^2 x so far I have done this: /sin x cos x\ sin x |----- + -----| ----- = sec^2 x \cos x sin x/ cos x /sin[x] sin[x] cos[x] cos[x]\ sin[x] |------ ------ + ------ ------| ------ = sec^2 x \sin[x] cos[x] cos[x] sin[x]/ cos[x] After that, I don't know how to work out this problem. I can't verify the identity. My second problem is: (1+cos x)(1-cos x) = sin^2 x So far I have done this: ( ? )(sin^2 x = 1-cos x) = sin^2 x I can't figure the rest out. Please help! Thank you! Baine
Date: 01/07/97 at 08:11:05 From: Doctor Pete Subject: Re: analytic trigonometry Hi, The first step is fine, but in the second step, I don't understand why you multiplied things by Sin[x]/Sin[x] and Cos[x]/Cos[x]. In order to get a clearer picture, it is often better not to immediately convert to sines and cosines. To see what's going on, let's make a few observations: Sin[x]^2 + Cos[x]^2 = 1 This is our familiar, basic trigonometric identity. We can derive many other identities from this; one of which arises from dividing both sides by Cos[x]^2: Sin[x]^2 Cos[x]^2 1 -------- + -------- = -------- Cos[x]^2 Cos[x]^2 Cos[x]^2 and since Tan[x] = Sin[x]/Cos[x], and Sec[x] = 1/Cos[x], this gives: Tan[x]^2 + 1 = Sec[x]^2 Now, to solve your problem, notice that: (Tan[x] + Cot[x])Tan[x] = Tan[x]^2 + Cot[x]Tan[x] But Cot[x] = 1/Tan[x]; that is, their product is 1. So the lefthand side of your equation is simply Tan[x]^2 + 1 And since we also showed that this was equal to Sec[x]^2, we have proved the identity. Notice that this is much cleaner than trying to convert into sines and cosines, because if we convert first and simplify later, what happens is that we no longer use the more "advanced" identities such as Tan[x]^2 + 1 = Sec[x]^2, because we are instead relying on the more basic identities between sine and cosine. This also makes it harder, because the more sophisticated identities are often proved using the more elementary ones, and so we have to incorporate these steps in between. In the second problem, notice that for some number n: (1+n)(1-n) = 1 - n + n - n^2 = 1 - n^2 So if n = Cos[x], then; (1+Cos[x])(1-Cos[x]) = 1 - Cos[x]^2 Since Cos[x]^2 + Sin[x]^2 = 1, the result follows immediately. Now, looking from the way you've been approaching these problems, it seems you need some help with understanding how to go about proving these identities. Generally, the idea is to start on one side of the equation, and work on that side to simplify it into the other side. For example, in the first problem, I worked only on the lefthand side, because it was more complicated and I readily saw something I could do with it. I used the distributive property, multiplying the Tan[x] through the inside. Now, you could have continued from the second step, as you wrote it above, but it is a more roundabout approach: / Sin[x] Sin[x] Cos[x] Cos[x] \ Sin[x] | ------ ------ + ------ ------ | ------ \ Sin[x] Cos[x] Cos[x] Sin[x] / Cos[x] / Sin[x]^2 + Cos[x]^2 \ Sin[x] = | ------------------- | ------ \ Sin[x]Cos[x] / Cos[x] / 1 \ Sin[x] = | ------------ | ------ \ Sin[x]Cos[x] / Cos[x] Sin[x] = --------------- Sin[x] Cos[x]^2 1 = -------- Cos[x]^2 = Sec[x]^2 What was missing was the observation that multiplying by Sin[x]/Sin[x] and Cos[x]/Cos[x] makes the denominators the same, so they can be combined. This makes the numerator Sin[x]^2 + Cos[x]^2 = 1, and the rest is straightfoward. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 01/07/97 at 17:24:07 From: Doctor Ceeks Subject: Re: analytic trigonometry Hi, I'd like to offer you an alternative answer to Doctor Pete's (which has its merits). Math is a very creative field and there are many ways to approach a problem, after all. In principle, there is only one trigonometric function from which all others can be derived, namely sin x (or cos x, if you prefer). As a matter of convenience, both sin x and cos x are used however, because it's a little easier to write cos x instead of sin (90-x) all the time. The fundamental relation is: sin^2 x + cos^2 x = 1 which is another way of writing the Pythagorean theorem. So your idea of converting everything to sine and cosine is fine. Now you were going fine. Multiplying by sin x/sin x and cos x/cos x as you did was a good way of making a common denominator for the two fractions in the sum tan x + cot x. Now you just have to go ahead and combine the fractions: (sin^2 x + cos^2 x) sin x ------------------- ----- sin x cos x cos x Using the fundamental identity, this becomes: 1 sin x ------------------- ----- sin x cos x cos x Cancelling like terms, we are left with: 1 ------- cos^2 x which is just sec^2 x, as desired. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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