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Math Help - identities

  1. #1
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    identities

    identities
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  2. #2
    Moo
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    Hello,

    Well, do you know identities for cos(a+b) and cos(a-b) ?

    cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

    So what will cos(a-b) be ?


    For the second one, use twice the formula for cos(a+b) :

    cos(3x)=cos(2x+x)=...
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  3. #3
    o_O
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    \cos(A + B) = {\color{red}\cos A \cos B} - \sin A \sin B
    \cos(A - B) = {\color{red} \cos A \cos B} + \sin A \sin B

    Add the equations together.

    If A = B:
    \cos (2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2} A - \sin^{2} A

    Convert into terms of only cosA and you should get your answer

    ---------

    Use your result from the first question to split your products in to sums. It'll be easier to integrate.

    ---------

    x = \cos t \: \: \Rightarrow \: \: dx = -\sin t dt

    Make the subs:

    \int \frac{\cos^{2} t}{\sqrt{1 - \cos^{2} t}} \: \cdot -\sin t dt \: \: = \: \:- \int \frac{\cos^{2} t \sin t}{\sqrt{\sin^{2} t}} dt

    etc.

    You'll need your half-angle identity from your first question to do it after you simplify.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by carlasader View Post
    identities
    For the third part \int\frac{x^2}{\sqrt{1-x^2}}

    let x=\cos(\theta)...dont often see this sub...
    then dx=-\sin(\theta)
    and x^2=\cos^2(\theta)

    we need to find the limits of integration so \frac{1}{2}=\cos(\theta)\Rightarrow(\theta=\frac{\  pi}{3}

    and 0=\cos(\theta)\Rightarrow{\theta=\frac{\pi}{2}}

    So putting that in we have

    \int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2(\  theta)\cdot{-\sin(\theta)}d\theta}{\sqrt{1-\cos^2(\theta)}} =\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\frac{\cos^2(  \theta)-\sin(\theta)d\theta}{\sin(\theta)}

    Simplifying we get -\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}\cos^2(\theta)  d\theta

    Using the identity \cos^2(\theta)=\frac{1+\cos(2\theta)}{2}

    we see this is equal to -\bigg[\frac{1}{2}\sin(2\theta)+\frac{\theta}{2}\bigg|_{\  frac{\pi}{2}}^{\frac{\pi}{3}}=.045293
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    \cos(A + B) = {\color{red}\cos A \cos B} - \sin A \sin B
    \cos(A - B) = {\color{red} \cos A \cos B} + \sin A \sin B

    Add the equations together.

    If A = B:
    \cos (2A) = \cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^{2} A - \sin^{2} A

    Convert into terms of only cosA and you should get your answer

    ---------

    Use your result from the first question to split your products in to sums. It'll be easier to integrate.

    ---------

    x = \cos t \: \: \Rightarrow \: \: dx = -\sin t dt

    Make the subs:

    \int \frac{\cos^{2} t}{\sqrt{1 - \cos^{2} t}} \: \cdot -\sin t dt \: \: = \: \:- \int \frac{\cos^{2} t \sin t}{\sqrt{\sin^{2} t}} dt

    etc.

    You'll need your half-angle identity from your first question to do it after you simplify.
    Darn you o_O! Well...its 1-2..but I think it should be 1.5-2 since I put the details ....
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