Trig Identites

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January 5th 2008, 08:13 AM
mathgeek777

Trig Identites

Hey guys.

In the following problem, I'm supposed to perform the addition and then use the fundamental identities to simplify.

Also please be loose on the format of the problem. I'm not very familiar with LaTeX at this point in time.

(1/(1 + cos x)) + (1/(1 - cos x))

What I did, since they had different denominators, I multiplied every fraction with (1 + cos x)(1 - cos x), eliminating the fractions and leaving me with

(1 - cos x) + (1 + cos x)

Thing is though, the answer is $2csc^2x$ , and the only thing I can do to the problem as I have it now is cancel the cos x and leave me with 2, which is obviously wrong. How is my thinking flawed on this problem?

I appreciate any and all help contributed.
January 5th 2008, 08:28 AM
Soroban

Hello, mathgeek777!

Quote:

Simplify: . $\frac{1}{1 + \cos x} + \frac{1}{1 - \cos x}$

You can't "eliminate the denominators".

We can get a common denominator, though . . .

$\frac{1}{1+\cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} + \frac{1}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}} \;\;=\; \;\frac{(1-\cos x) + (1 + \cos x)}{(1+\cos x)(1-\cos x)}$

. . $= \;\;\frac{2}{1-\cos^2\!x} \;\;=\;\;\frac{2}{\sin^2\!x} \;\;=\;\;2\csc^2\!x$
January 5th 2008, 08:44 AM
topsquark

Quote:

Originally Posted by mathgeek777

Hey guys.

In the following problem, I'm supposed to perform the addition and then use the fundamental identities to simplify.

Also please be loose on the format of the problem. I'm not very familiar with LaTeX at this point in time.

(1/(1 + cos x)) + (1/(1 - cos x))

What I did, since they had different denominators, I multiplied every fraction with (1 + cos x)(1 - cos x), eliminating the fractions and leaving me with

(1 - cos x) + (1 + cos x)

Thing is though, the answer is $2csc^2x$ , and the only thing I can do to the problem as I have it now is cancel the cos x and leave me with 2, which is obviously wrong. How is my thinking flawed on this problem?

I appreciate any and all help contributed.

You are simplifying this as if it is an equation where you can multiply both sides by something and come up with an equivalent equation. But this isn't an equation so you can't do the multiplication.

-Dan
January 5th 2008, 09:23 AM
mathgeek777

Quote:

Originally Posted by Soroban

Hello, mathgeek777!

You can't "eliminate the denominators".

We can get a common denominator, though . . .

$\frac{1}{1+\cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} + \frac{1}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1+\cos x}} \;\;=\; \;\frac{(1-\cos x) + (1 + \cos x)}{(1+\cos x)(1-\cos x)}$

. . $= \;\;\frac{2}{1-\cos^2\!x} \;\;=\;\;\frac{2}{\sin^2\!x} \;\;=\;\;2\csc^2\!x$

...Oh! I knew that. Thanks! (Handshake) I appreciate your assistance.

I do have another question though. How did you go from $\;\frac{(1-\cos x) + (1 + \cos x)}{(1+\cos x)(1-\cos x)}$ to $\frac{2}{1-\cos^2\!x}$

EDIT: Forgive the brain fart, I just realized how you did that.