Verifying Trig Identities

**kelsey3** · Apr 6th 2008, 02:48 PM

1. Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx
2. Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1

**Soroban** · Apr 6th 2008, 04:58 PM

Hello, kelsey3!

2. Verify the identity: . $\frac{1}{1-\cos x} - \frac{\cos x}{1+\cos x} \;=\;2\csc^2\!x-1$

On the left side, get a common denominator and combine . . .

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

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

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

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

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