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### 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.

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/

Associated Topics:
High School Trigonometry

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