# Verifying Trigonometric Identity

• May 16th 2008, 09:08 AM
mathgeek777
Verifying Trigonometric Identity
Hello.

I'm currently doing chapter reviews for my final exam, and I've come across a trigonometric identity that I have seen before, but don't recall how to verfiy. If you can give me any help and explain it, it will be greatly appreciated. The identity is below.

$\displaystyle \sqrt{\frac{1-sin\theta}{1+sin\theta}}=\frac{1-sin\theta}{|cos\theta|}$

Bryan
• May 16th 2008, 09:16 AM
flyingsquirrel
Hi

Hint : $\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}$
• May 16th 2008, 09:19 AM
mathgeek777
Quote:

Originally Posted by flyingsquirrel
Hi

Hint : $\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}$

Yeah, my gut was saying to multiply the radical by its conjugate, but I'm really not sure how to get rid of the radical itself. Not to mention I dont know where $\displaystyle |\cos\theta|$ comes from.
• May 16th 2008, 09:25 AM
Moo
Hello,

Quote:

Originally Posted by mathgeek777
Yeah, my gut was saying to multiply the radical by its conjugate, but I'm really not sure how to get rid of the radical itself. Not to mention I dont know where $\displaystyle |\cos\theta|$ comes from.

$\displaystyle 1-\sin^2x=\cos^2x$ (this is the second hint for the denominator :p)
• May 16th 2008, 09:57 AM
abender
$\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}} =\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{1-\sin^2}}$
$\displaystyle =\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{\cos^2 \theta}}=\frac{\sqrt{(1-\sin\theta)(1-\sin\theta)}}{\sqrt{\cos^2\theta}} =\frac{1-\sin\theta}{|\cos^2\theta|}$