# Trig identities!!

• Nov 5th 2006, 06:35 PM
xxpl510xx
Trig identities!!
Hi all, Im stumped on this one:

(sin s)/(1 + cos s) + (1 + cos s)/(sins s) = 2csc s :confused:

Thanks for you help.
• Nov 5th 2006, 06:39 PM
ThePerfectHacker
Quote:

Originally Posted by xxpl510xx
Hi all, Im stumped on this one:

(sin s)/(1 + cos s) + (1 + cos s)/(sins s) = 2csc s :confused:

Thanks for you help.

$\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}$

$\frac{(\sin x)(\sin x)+(1+\cos x)(1+\cos x)}{\sin x(1+\cos x)}$

$\frac{\sin^2 x+1+2\cos x+\cos^2 x}{\sin x(1+\cos x)}$

$\frac{1+1+2\cos x}{\sin x(1+\cos x)}$

$\frac{2(1+\cos x)}{\sin x(1+\cos x)}$

$\frac{2}{\sin x}$

$2\csc x$
• Nov 6th 2006, 06:12 AM
Soroban
Hello, xxpl510xx!

Another approach . . .

Quote:

$\frac{\sin \theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} \:= \:2\csc\theta$

Multiply the first fraction by $\frac{1-\cos\theta}{1 - \cos\theta}$

. . $\frac{\sin\theta}{1 + \cos\theta}\cdot\frac{1 - \cos\theta}{1 - \cos\theta} \:=\:\frac{\sin\theta(1 - \cos\theta)}{1-\cos^2\theta} \:=\:\frac{\sin\theta(1 - \cos\theta)}{\sin^2\theta}$ $=\:\frac{1 - \cos\theta}{\sin\theta}$

And the problem becomes: . $\frac{1-\cos\theta}{\sin\theta} + \frac{1 + \cos\theta}{\sin\theta} \:=\:\frac{2}{\sin\theta} \:=\:2\csc\theta$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is a popular "trick" to use when dealing with:
. . . $1 + \sin\theta,\;1 - \sin\theta,\;1 + \cos\theta,$ or $1 - \cos\theta$

It arises from: . $\sin^2\theta + \cos^2\theta\:=\:1\quad\Rightarrow\quad \sin^2\theta \:=\:1-\cos^2\theta$

. . $\Rightarrow\quad(\sin\theta)(\sin\theta) \:=\:(1-\cos\theta)(1 + \cos\theta)$

which produces variations like: . $\frac{\sin\theta}{1 + \cos\theta} \:=\:\frac{1 - \cos\theta}{\sin\theta}$

It also applies to: . $\sec\theta +1$ and $\sec\theta - 1$

Just one more thing to watch for . . .