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### Simplifying Trigonometric Equations

```
Date: 06/02/98 at 10:13:08
From: Chris Keenan
Subject: Trigonometric equations

I am a student at Portsmouth high school in NH and our math teacher
asked us to answer the following problems with help from the Internet.
We found you, so here goes. (He didn't want us to try it ourselves.)

Problem 1:

cos 2x            cos 2x
--------------- - --------------- - csc 2x
sec 2x - tan 2x   sec 2x + tan 2x

Problem 2:

sin x
tan(x/2) = ---------     where 0 < x < (Pi/2)
1 + cos x

I hope you can help. Thanks.

Chris
```

```
Date: 06/02/98 at 12:57:16
From: Doctor Barrus
Subject: Re: Trigonometric equations

Hi, Chris!

Well, these look like some interesting problems. Since it looks like
none of the other math doctors has answered you yet, I guess I'll give
it a shot.

Problem 1:

I'm assuming that the problem is to simplify this as much as possible.
The first thing I usually do in problems like this is to write
everything in terms of sines and cosines. I do this because I know and
can derive more identities involving sines and cosines, and textbooks
give more identities for sines and cosines than for other
trigonometric functions. So let's do that:

cos 2x                 cos 2x              1
------------------- - --------------------- - --------
1       sin 2x         1        sin 2x      sin 2x
------ - --------     -------- + --------
cos 2x    cos 2x       cos 2x     cos 2x

Now I'll simplify the denominators:

cos 2x              cos 2x          1
---------------- - ---------------- - --------
1 - sin 2x         1 + sin 2x       sin 2x
------------       ------------
cos 2x             cos 2x

and simplify the fraction division:

(cos 2x)^2     (cos 2x)^2       1         (^2 means raised to
------------ - ------------ - --------      the second power)
1 - sin 2x     1 + sin 2x     sin 2x

Next I'll get a common denominator for the first two terms, and add:

(1 + sin 2x)(cos 2x)^2 - (1 - sin 2x)(cos 2x)^2       1
------------------------------------------------- - --------
(1 - sin 2x)(1 + sin 2x)                 sin 2x

Then I'll simplify the expressions in the numerator and denominator of
the first term a bit:

[1 + sin 2x - (1 - sin 2x)](cos 2x)^2       1
--------------------------------------- - --------
1 - (sin 2x)^2                  sin 2x

and then a little more:

(2 sin 2x)(cos 2x)^2       1        (Here I used the identity
---------------------- - --------     (sin x)^2 + (cos x)^2 = 1)
(cos 2x)^2          sin 2x

and some more

1
2 sin 2x - --------
sin 2x

Now I have something that really looks good. I'll put everything over
a common denominator and subtract:

2(sin 2x)^2 - 1
-----------------
sin 2x

Looking at this, I'd say that this looks similar to something in
the identity:

cos 2k = 1 - 2(sin k)^2

If I put 2x in everywhere there's an k, and multiply by -1, here's
what I get:

-cos 4x = 2(sin 2x)^2 - 1

So now I can reduce what I have to:

cos 4x
- --------
sin 2x

This looks a ton better than what we started out with, doesn't it? At
this point I would say that this is the simplest form I can write this
in. Of course, as with most trigonometric expressions, there are lots
of ways to write this in other forms. For example, we could write this
particular expression as -(cos 4x)(csc 2x), so the exact form of your
answer is going to be somewhat a matter of personal taste. Since I
like sines and cosines, I'm going to keep my answer as it is. So

cos 2x            cos 2x                    cos 4x
--------------- - --------------- - csc 2x = - --------
sec 2x - tan 2x   sec 2x + tan 2x               sin 2x

Okay. Now on to the next one:

Problem 2:

textbook:

sin(x/2) = sqrt[(1 - cos x)/2]
cos(x/2) = sqrt[(1 + cos x)/2]
tan x = (sin x)/(cos x)

where sqrt means the square root.

Then:

sin(x/2)
tan(x/2) = ----------
cos(x/2)

sqrt[(1 - cos x)/2]
= ---------------------
sqrt[(1 + cos x)/2]

sqrt(1 - cos x)/sqrt(2)   (A property of square
= -------------------------   roots of real numbers.)
sqrt(1 + cos x)/sqrt(2)

sqrt(1 - cos x)
= ----------------  (Simplifying the fraction)
sqrt(1 + cos x)

[sqrt(1 - cos x)][sqrt(1 + cos x)]
= ------------------------------------
[sqrt(1 + cos x)][sqrt(1 + cos x)]

because multiplying the top and bottom of a fraction by the same
nonzero number does not change the fraction value (and we know that
the value is nonzero, because x < pi).

sqrt[(1 - cos x)(1 + cos x)]
= ------------------------------  (Properties of square
1 + cos x                 roots.)

sqrt[1 - (cos x)^2]
= ---------------------
1 + cos x

sqrt[(sin x)^2]     (because (sin x)^2 + (cos x)^2 = 1)
= -----------------
1 + cos x

sin x
= -----------
1 + cos x

as desired. Please note that it is important to keep 0 < x < pi. When
x is in this range, both sin x and cos x have positive values, so the
rules I have used for square roots work.

Well, this has been a long answer, but I hope it helps. Good luck!

-Doctor Barrus,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 06/02/98 at 13:41:39
From: Doctor Barrus
Subject: Re: Trigonometric equations

By the way, Chris, a lot of people have been mailing this question in
(I assume they're your classmates). Their questions have been a little
different, though. Instead of having the first thing as cos 2x, they
have cot 2x. You might want to check the assignment. Just in case you
mistyped it, I've modified my answer to fit the other problem.
Here it is:

cot 2x            cos 2x
--------------- - --------------- - csc 2x
sec 2x - tan 2x   sec 2x + tan 2x

I'm assuming that the problem is to simplify this as much as possible.
The first thing I usually do in problems like this is to write
everything in terms of sines and cosines. I do this because I know and
can derive more identities involving sines and cosines, and textbooks
give more identities for sines and cosines, than for other
trigonometric functions. So let's do that:

cos 2x
--------
sin 2x                 cos 2x              1
------------------- - --------------------- - --------
1       sin 2x         1        sin 2x      sin 2x
------ - --------     -------- + --------
cos 2x    cos 2x       cos 2x     cos 2x

Now I'll simplify the denominators:

cos 2x
--------
sin 2x              cos 2x          1
---------------- - ---------------- - --------
1 - sin 2x         1 + sin 2x       sin 2x
------------       ------------
cos 2x             cos 2x

and simplify the fraction division:

(cos 2x)^2          (cos 2x)^2       1
-------------------- - ------------ - --------
(sin 2x)(1 - sin 2x)    1 + sin 2x     sin 2x

where ^2 means raised to the second power. Next I'll get a common

(1 + sin 2x)(cos 2x)^2 - (sin 2x)(1 - sin 2x)(cos 2x)^2
- (1 - sin 2x)(1 + sin 2x)
---------------------------------------------------------
(sin 2x)(1 - sin 2x)(1 + sin 2x)

Then I'll simplify the expressions in the numerator and denominator of
the first term a bit:

[1 + sin2x - (sin2x)(1 - sin2x)](cos2x)^2 - [1 - (sin2x)^2]
-----------------------------------------------------------
(sin 2x)(1 - (sin 2x)^2)

and then a little more (using the identity (sin x)^2 + (cos x)^2 = 1):

[1 + (sin 2x)^2](cos 2x)^2 - (cos 2x)^2
---------------------------------------
(sin 2x)(cos 2x)^2

and some more:

(sin 2x)^2
------------
sin 2x

And finally divide:

sin 2x

This looks a lot better than what we started out with, doesn't it? At
this point I would say that this is the simplest form I can write this
in. Of course, as with most trigonometric expressions, there are many
ways to write this in other forms. For example, we could write this
particular expression as 2(sin x)(cos x), so the exact form of your
answer is going to be somewhat a matter of personal taste. Since it's
pretty compact, I'm going to keep my answer as it is. So

cot 2x            cos 2x
--------------- - --------------- - csc 2x  = sin 2x
sec 2x - tan 2x   sec 2x + tan 2x

Good luck!

-Doctor Barrus,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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