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Math Help - trig identites, help needed bad

  1. #1
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    trig identites, help needed bad

    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
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  2. #2
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    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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