The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

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!


Date: 01/07/97 at 08:11:05
From: Doctor Pete
Subject: Re: analytic trigonometry


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 

     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 

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] 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!   

Date: 01/07/97 at 17:24:07
From: Doctor Ceeks
Subject: Re: analytic trigonometry


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 

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:

cos^2 x

which is just sec^2 x, as desired.

-Doctor Ceeks,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.