# Verifying Trig Identities

• Apr 6th 2008, 02:48 PM
kelsey3
Verifying Trig Identities
1. Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx
2. Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1
• Apr 6th 2008, 04:58 PM
Soroban
Hello, kelsey3!

Quote:

2. Verify the identity: .$\displaystyle \frac{1}{1-\cos x} - \frac{\cos x}{1+\cos x} \;=\;2\csc^2\!x-1$
On the left side, get a common denominator and combine . . .

$\displaystyle \frac{1}{1-\cos x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} - \frac{\cos x}{1+\cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;\;=\;\;\frac{1+\cos x}{1-\cos^2\!x} - \frac{\cos x(1 - \cos x)}{1-\cos^2\!x}$

. . $\displaystyle = \;\;\frac{1 + \cos x - \cos x(1 - \cos x)}{1-\cos^2\!x} \;\;=\;\;\frac{1 + \cos x - \cos x + \cos^2\!x}{1-\cos^2\!x}$

. . $\displaystyle = \;\;\frac{1 + \cos^2\!x}{\sin^2\!x} \;\;=\;\;\frac{1 + (1 - \sin^2\!x)}{\sin^2\!x} \;\;=\;\;\frac{2 - \sin^2\!x}{\sin^2\!x}$

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