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Math Help - Trigonometry Identities Proof

  1. #1
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    Trigonometry Identities Proof

    I really need help with this problem.
    sin2X+sin4X tan3X
    __________ + _____ =0
    sin2X-sin4X tanX

    they are 2 fractions added together. It all equals 0. PLEASE help!
    the 2X and 4X are not exponents they are identites, for trig functions. Thanks!
    Last edited by omgilymcr; April 4th 2012 at 02:44 PM.
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  2. #2
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    Re: Trigonometry Identities Proof

    Hello, omgilymcr!

    We will need these identities:

    . . \sin2\theta \:=\:2\sin\theta\cos\theta

    . . \sin3\theta \:=\:3\sin\theta - 4\sin^3\theta

    . . \cos3\theta \:=\:4\cos^3\theta - 3\cos\theta

    . . \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}



    \text{Verify: }\:\frac{\sin2x + \sin4x}{\sin2x-\sin4x} + \frac{\tan3x}{\tan x} \:=\:0

    The first fraction is: . \frac{\sin 2x + \sin4x}{\sin2x - \sin4x} \;=\;\frac{\sin2x + 2\sin2x\cos2x}{\sin2x - 2\sin2x\cos2x}

    . . . . . . . . . . =\;\frac{\sin2x(1 + 2\cos2x)}{\sin2x(1 - 2\cos2x)} \;=\;\frac{1+2\cos2x}{1-2\cos2x}\;\;{\color{blue}[1]}


    The second fraction is: . \frac{\tan3x}{\tan x} \;=\;\frac{\sin3x}{\cos3x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3\sin x - 4\sin^3x}{4\cos^3x - 3\cos x}\cdot\frac{\cos x}{\sin x}

    . . . . . . . . . . =\;\frac{\sin x(3 - 4\sin^2x)}{\cos x(4\cos^2x - 3)}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3-4\sin^2x}{4\cos^2x-3}

    . . . . . . . . . . =\;\frac{3-4(1-\cos^2x)}{4\cos^22x-3} \;=\;\frac{4\cos^2x-1}{4\cos^2x-3}

    . . . . . . . . . . =\;\frac{4\left(\frac{1+\cos^2x}{2}\right) - 1}{4\left(\frac{1+\cos^2x}{2}\right)-3} \;=\;\frac{1 + 2\cos2x}{2\cos2x-1}

    . . . . . . . . . . =\;-\frac{1 + 2\cos2x}{1 - 2\cos2x}\;\;{\color{blue}[2]}


    We see that [2] is the negative of [1].

    Therefore, their sum is zero.


    Edit: corrected typo.
    .
    Last edited by Soroban; April 5th 2012 at 04:55 AM.
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  3. #3
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    Re: Trigonometry Identities Proof

    Quote Originally Posted by Soroban View Post
    Hello, omgilymcr!

    We will need these identities:

    . . \sin2\theta \:=\:2\sin\theta\cos\theta

    . . \sin3\theta \:=\:3\sin\theta - 4\sin^3\theta

    . . \cos3\theta \:=\:4\cos^3\theta - 3\cos\theta

    . . \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}




    The first fraction is: . \frac{\sin 2x + \sin4x}{\sin2x - \sin4x} \;=\;\frac{\sin2x + 2\sin2x\cos2x}{\sin2x - 2\sin2x\cos2x}

    . . . . . . . . . . =\;\frac{\sin2x(1 + 2\cos2x)}{\sin2x(1 - 2\cos2x)} \;=\;\frac{1+2\cos2x}{1-2\cos2x}\;\;{\color{blue}[1]}


    The second fraction is: . \frac{\tan3x}{\tan x} \;=\;\frac{\sin3x}{\cos3x}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3\sin x - 4\sin^3x}{4\cos^3x - 3\cos x}\cdot\frac{\cos x}{\sin x}

    . . . . . . . . . . =\;\frac{\sin x(3 - 4\sin^2x)}{\cos x(4\cos^2x - 3)}\cdot\frac{\cos x}{\sin x} \;=\;\frac{3-4\sin^2x}{4\cos^2x-3}

    . . . . . . . . . . =\;\frac{3-4(1-\cos^2x)}{4\cos^22x-3} \;=\;\frac{4\cos^2x-1}{4\cos^2x-3}

    . . . . . . . . . . =\;\frac{4\left(\frac{1+\cos^2x}{2}\right) - 1}{4\left(\frac{1+\cos^2x}{2}\right)-3} \;=\;\frac{1 + 2\cos2x}{2\cos2x-1}

    . . . . . . . . . . =\;-\frac{1 + 2\cos2x}{1 - \cos2x}\;\;{\color{blue}[2]}


    We see that [2] is the negative of [1].

    Therefore, their sum is zero.

    You are missing a 2 in the denominator of [2]...
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