Trigonometric Identities

**kelsey3** · Apr 6th 2008, 02:28 PM

Help!
1. Verify the identity: (1/1-cosx)-(cosx/1+cosx)=2csc(squared)x-1
2.Verify the identity: cos5t-cos3t= -8sin(squared)t(2cos(cubed)t-cost)
3. Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx

**galactus** · Apr 6th 2008, 02:44 PM

#1:

$\frac{1}{1-cosx}-\frac{cosx}{1+cosx}=2csc^{2}x-1$

The left side, you can cross multiply into:

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

$=\frac{1+cos^{2}x}{1-cos^{2}x}$

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

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

$=csc^{2}x+cot^{2}x$

Use the identity: $csc^{2}x-1=cot^{2}x$

$csc^{2}x+(csc^{2}x-1)=2csc^{2}x-1$

**kelsey3** · Apr 6th 2008, 04:23 PM

Do you know how to do the other two?....lol

**Soroban** · Apr 6th 2008, 05:16 PM

Hello, kelsey3!

We need these double-angle identities:
. . $\sin2\theta \:=\:2\sin\theta\cos\theta$
. . $\cos2\theta \:=\:2\cos^2\!\theta-1$

And this sum-to-product identity: . $\cos A - \cos B \:=\:-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

2. Verify the identity: . $\cos5t-\cos3t\l= \; -8\sin^2\!t\cos t\left(2\cos^2\!t-1\right)$

The left side is: . $-2\sin\left(\frac{5t+3t}{2}\right)\sin\left(\frac{5 t-3t}{2}\right)$

. . . . . . . . . . $=\;-2\sin4t\sin t$

. . . . . . . . . . $= \;-2(2\sin2t\cos2t) \sin t$

. . . . . . . . . . $=\;-4\sin2t\cos2t\sin t$

. . . . . . . . . . $= \;-4(2\sin t\cos t)(2\cos^2\!t-1)\sin t$

. . . . . . . . . . $= \;-8\sin^2\!t\cos t(2\cos^2\!t - 1)$

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