trig identites, help needed bad

**mehadi91kochotmail** · November 6th 2008, 04:18 PM

1) sin8x= 8SinxCosxCox2xCos4x
2) sin^4x + cos^4x = sin^2x(csc^2x-2cos^2x)
3) sin4x/1-cos4x X 1-cos2x/cos2x= tan x
4) sin2x/1+cos2x X cosx/1+cosx= tan x/2

if anyone could prove any of these it will help alot

**Soroban** · November 6th 2008, 05:14 PM

Hello, mehadi91kochotmail!

I'll do a couple of these for you . . .

$2)\;\;\sin^4\!x + \cos^4\!x \:= \:\sin^2\!x(\csc^2\!x - 2\cos^2\!x)$

This one is really goofy . . .

On the left side, add and subtract $2\sin^2\!x\cos^2\!x$

. . $\underbrace{\sin^4\!x \:{\color{blue}+\: 2\sin^2\!x\cos^2\!x} + \cos^4\!x} \:{\color{blue}- \:2\sin^2\!x\cos^2\!x}$

. . . . $= \;(\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}})^2 - 2\sin^2\!x\cos^2\!x$

. . . . $= \;1 - 2\sin^2\!x\cos^2\!x$

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

Factor: . $\sin^2\!x\left[\frac{1}{\sin^2\!x} - 2\cos^2\!x\right]$

. . . . $= \;\sin^2\!x\left(\csc^2\!x - 2\cos^2\!x\right)$

$4)\;\;\frac{\sin2x}{1+\cos2x}\cdot \frac{\cos x}{1+ \cos x} \:=\: \tan\tfrac{x}{2}$

For this one, we need more identities:

. . $\sin2\theta \:=\:2\sin\theta\cos\theta \qquad\qquad 1 + \cos2\theta \:=\:2\cos^2\theta$

Left side: . $\frac{\sin2x}{1 + \cos2x}\cdot\frac{\cos x}{1 + \cos x} \:=\:\frac{2\sin x\cos x}{2\cos^2\!x}\cdot\frac{\cos x}{1 + \cos x} \;=\;\frac{\sin x}{1 + \cos x}$

. . . . . $= \;\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\! \frac{x}{2}} \;=\;\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \;=\;\tan\tfrac{x}{2}$

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