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Math Help - Challenging Trig Identity Questions

  1. #1
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    Challenging Trig Identity Questions

    Need some help with

    (9)\;\;1 - \frac{\sin^2\!x\tan x}{\tan x+1} - \frac{\cos^2\!x}{\tan x+1} \;=\;\sin x\cos x

    Thanks.


    Moderator edit: Scanned images deleted, single question typed out.
    Last edited by mr fantastic; December 3rd 2011 at 12:52 PM.
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  2. #2
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    Re: Challenging Trig Identity Questions

    Hello, tmac20!

    Can you show us any of your work?

    (9)\;\;1 - \frac{\sin^2\!x\tan x}{\tan x+1} - \frac{\cos^2\!x}{\tan x+1} \;=\;\sin x\cos x
    We have: . 1 - \dfrac{\dfrac{\sin^2\!x}{1}\,\dfrac{\sin x}{\cos x}}{\dfrac{\sin x}{\cos x} + 1} - \dfrac{\cos^2\!x}{\dfrac{\sin x}{\cos x} + 1}


    Multiply both fractions by \frac{\cos x}{\cos x}\!:\;\;1 - \frac{\sin^3\!x}{\sin x + \cos x} - \frac{\cos^3\!x}{\sin x+\cos x}

    . . =\;\;1 - \frac{\overbrace{\sin^3\!x + \cos^3\!x}^{\text{sum of cubes}}}{\sin x+\cos x} \;\;=\;\;1 - \frac{(\sin x+\cos x)(\sin^2\!x - \sin x\cos x + \cos^2x)}{\sin x+\cos x}


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

    . . =\;\;1 - 1 + \sin x\cos x \;\;=\;\;\sin x\cos x
    Last edited by mr fantastic; December 3rd 2011 at 12:52 PM.
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