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Math Help - Differentiation's Proof ..

  1. #1
    Ted
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    Differentiation's Proof ..

    Show that:
     \frac{d}{dx} \left( \frac{sin^2(x)}{1+cot(x)} + \frac{cos^2(x)}{1+tan(x)} \right) = -cos(2x).

    Am sure, there is a trick somewhere.
    sure, we will not differentiate it normally.
    I rewrite the function as:
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  2. #2
    MHF Contributor Calculus26's Avatar
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    the function reduces to 1- 1/2sin(2x)

    convert to sines and cosines and combine into a single term

    recall a^3 + b^3 = (a+b) (a^2 -ab +b^2)
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  3. #3
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    Hello, Ted!

    You're right . . . We're expected to simplify the function first.


    \text{Show that: }\;\frac{d}{dx}\left( \frac{\sin^2x}{1+\cot x} + \frac{\cos^2x}{1+\tan x } \right) \;=\; -\cos 2x.

    We have: . \frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}

    . . . . . . =\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} \;+\; {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}

    . . . . . . =\;\frac{\sin^3\!x}{\sin x + \cos x} + \frac{\cos^3\!x}{\sin x + \cos x}

    . . . . . . =\;\frac{\sin^3\!x + \cos^3\!x}{\sin x + \cos x}

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

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

    . . . . . . =\; 1 - \sin x\cos x

    . . . . . . =\;1 - \tfrac{1}{2}\sin2x


    Now differentiate it!



    Edit: Ah, Calculus 26 beat me to it . . . *sigh*
    .
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Ted!

    You're right . . . We're expected to simplify the function first.


    We have: . \frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}

    . . . . . . =\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} \;+\; {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}

    . . . . . . =\;\frac{\sin^3\!x}{\sin x + \cos x} + \frac{\cos^3\!x}{\sin x + \cos x}

    . . . . . . =\;\frac{\sin^3\!x + \cos^3\!x}{\sin x + \cos x}

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

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

    . . . . . . =\; 1 - \sin x\cos x

    . . . . . . =\;1 - \tfrac{1}{2}\sin2x


    Now differentiate it!


    Edit: Ah, Calculus 26 beat me to it . . . *sigh*
    .
    the colors the indenting... \;\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is cool}} ...too much
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