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Math Help - Another Compound Trig Identity

  1. #1
    Newbie
    Joined
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    Another Compound Trig Identity

    I'm not sure if what I did was right, so please verify it.

    sin 2x/(1-cos 2x)=2 csc 2x-tan x

    This is what I did:

    LS= sin 2x/(1-cos2x)
    = (2sinx cosx)/(1-(cos^2 x-sin^2 x))
    = (2sinx cosx)/(cos^2 x+sin^2 x-cos^2 x+sin^2 x)
    = (2 sin x cos x)/(2sin^2 x)
    = cot x

    RS= 2/(2sinx cosx)-sinx/cosx
    = (2-2sin^2 x)/(2sinx cosx)
    = (1-sin^2 x)/(sinx cosx)
    = (1-(1-cos^2 x))/(sinx cosx)
    = cos^2 x/(sinx cosx)
    = cot x
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  2. #2
    Super Member

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    May 2006
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    Lexington, MA (USA)
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    Hello, DemonX01!

    Your work is correct.
    However, you should not work with both sides of the identity.

    I did find a way to change one side into the other . . .


    Verify: . \frac{\sin 2x}{1-\cos2x} \:=\:2\csc 2x-\tan x
    I started with the right side . . .

    2\csc2x - \tan x \;\;=\;\;\frac{2}{\sin2x} - \frac{\sin x}{\cos x} \;\;=\;\;\frac{2}{2\sin x\cos x} \;-\; \frac{\sin x}{\cos x}

    . . . . . =\;\;\frac{1}{\sin x\cos x} - \frac{\sin x}{\cos x} \;\;=\;\;\frac{1}{\sin x\cos x} - \frac{\sin x}{\cos x}\cdot{\color{blue}\frac{\sin x}{\sin x}}

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


    Multiply by \frac{2\sin x}{2\sin x}\!:\;\;\frac{2\sin x}{2\sin x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{2\sin x\cos x}{2\sin^2\!x} \;=\;\frac{\sin2x}{1-\cos 2x}\;\;\begin{array}{c} \\ {\color{red}**} \end{array}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    The denomniator comes from the identity:

    . . \sin^2\theta \:=\:\frac{1-\cos2\theta}{2} \quad\Rightarrow\quad 2\sin^2\!\theta \:=\:1 - \cos2\theta

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