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Math Help - Prove the following Trig Identity

  1. #1
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    Prove the following Trig Identity

    Sorry for troubling you guys but i keep coming up with an extra 3cosX

    Prove:

    3sinxtanx=8 is the same as 3cos^2x+8cosx-3=0


    Any help would be great.
    Last edited by Dominic O'L; April 2nd 2009 at 06:32 AM.
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  2. #2
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    Quote Originally Posted by Dominic O'L View Post
    Sorry for troubling you guys but i keep coming up with an extra 3cosX

    Prove:

    3sinxtanx=8 is the same as 3cos^2x+8cosx-3=0


    Any help would be great.
    tan{x} = \frac{sin(x)}{cos(x)}

    \frac{3sin^2(x)}{cos(x)}-8 = 0

    \frac{3-3cos^2(x)}{cos(x)}-8=0

    -8 = -8\frac{cos(x)}{cos(x)}

    \frac{3-3cos^2(x)}{cos(x)}-8\frac{cos(x)}{cos(x)}=0

    \frac{3-3cos^2(x)-8cos(x)}{cos(x)} = 0

    Then cos(x) will disappear and take everything to the other side to get the answer
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  3. #3
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    Quote Originally Posted by Dominic O'L View Post
    Sorry for troubling you guys but i keep coming up with an extra 3cosX

    Prove:

    3sinxtanx=8 is the same as3cos^2x+8cosx-3=0


    Any help would be great.
    Hi Dominic,

    I'm not sure if this is the proof you are looking for, but here is something for what it's worth.

    3 \cos^2x+8 \cos x -3=0

    (3 \cos x-1)(\cos x +3)=0

    3\cos x-1=0

    \cos x={\color{red}\frac{1}{3}}

    Now, use that in the 1st equation.

    3 \sin x \tan x = 8

    3 \sin x \left(\frac{\sin x}{\cos x}\right)=8

    \frac{3 \sin^2x}{\cos x}=8

    \frac{3 \sin^2 x}{{\color{red}\frac{1}{3}}}=8

    9 \sin^2x=8

    \sin^2x=\frac{8}{9}

    \sin x=\frac{2\sqrt{2}}{3}

    Therefore, \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)=\cos^{-1} \left(\frac{1}{3}\right)
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    Thank you ever so much, i'm in the process of working through maths past papers so your help is much appreciated, AS exam only a month and a half away.
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  5. #5
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    Hello, Dominic O'L

    Prove: . 3\sin x\tan x\:=\:8 is the same as 3\cos^2\!x+ 8\cos x -3\:=\:0

    We have: . 3\sin x\cdot\frac{\sin x}{\cos x} \:=\:8 \quad\Rightarrow\quad \frac{3\sin^2\!x}{\cos x} \:=\:8 \quad\Rightarrow\quad 3\sin^2\!x \:=\:8\cos x


    . . 3(1-\cos^2\!x) \:=\:8\cos x \quad\Rightarrow\quad 3-3\cos^2\!x \:=\: 8\cos x


    Therefore: . \boxed{3\cos^2\!x + 8\cos x - 3 \:=\:0}

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