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Math Help - cos2x

  1. #1
    Member srirahulan's Avatar
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    Arrow cos2x

    How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))
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  2. #2
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    Re: cos2x

    just expand out \frac{\cos\left(2(x+h)\right)-\cos(2x)}{h} and take the limit as h \to 0.
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  3. #3
    Member srirahulan's Avatar
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    Re: cos2x

    but i asked to use cos2x proofing by cosx... i tried this way,,, cos2x=2cos^2(x)-1, and then expand, finally i stacked up.... please give me a solution????
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  4. #4
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    Re: cos2x

    Quote Originally Posted by srirahulan View Post
    but i asked to use cos2x proofing by cosx... i tried this way,,, cos2x=2cos^2(x)-1, and then expand, finally i stacked up.... please give me a solution????
    sheesh

    \frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=

    \frac{\cos(2x+2h)-cos(2x)}{h}=-2\sin(2x)

    \frac{\cos(2x)\cos(2h)-\sin(2x)\sin(2h)-\cos(2x)}{h}=

    \frac{\cos(2x)(cos(2h)-1)-\sin(2x)\sin(2h)}{h}=

    \cos(2x)\frac{cos(2h)-1}{h}-\sin(2x)\frac{\sin(2h)}{h}

    we need to reduce the sin and cos terms that involve h further so we can use known limits

    \cos(2h)=2\cos^2(h)-1$ and $\sin(2h)=2\sin(h)\cos(h)

    so

    \displaystyle{\lim_{h\to 0}}\frac{\cos(2h)-1}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\cos^2(h)-2}{h}=0

    \displaystyle{\lim_{h\to 0}}\frac{\sin(2h)}{h}=\displaystyle{\lim_{h\to 0}}\frac{2\sin(h)\cos(h)}{h}=2

    and thus

    \displaystyle{\lim_{h\to 0}}\frac{\cos\left(2(x+h)\right)-\cos(2x)}{h}=-2\sin(2x)
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  5. #5
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    Re: cos2x

    Quote Originally Posted by srirahulan View Post
    How can i prove cos2x derivation using cos x by basic derivation proof.... like (limit and delta(x) delta(y))
    It depends on what you already know. Are you allowed to use the Chain Rule? If so, you can prove \displaystyle \begin{align*} \frac{d}{dx} \left[ \cos{(x)} \right] = -\sin{(x)} \end{align*} by first principles and then apply the chain rule. If you really wanted to, you could prove \displaystyle \begin{align*} \frac{d}{dx} \left[ \sin{(x)} \right] = \cos{(x)} \end{align*} by first principles, then remember \displaystyle \begin{align*} \cos{(x)} \equiv \sin{ \left( \frac{\pi}{2} - x \right) } \end{align*} and use the chain rule from there...
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    Re: cos2x

    cos2x-14-jan-14-2.png
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