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Math Help - An awkward cosine problem

  1. #1
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    An awkward cosine problem

    A student came to me with this problem, and sadly I do not know how to finish this problem, any help, please?

     \cos ^2 ( \frac { \pi }{8} - \frac {1}{2} )
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    What exactly do you want to do with this?
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    Quote Originally Posted by tttcomrader View Post
    A student came to me with this problem, and sadly I do not know how to finish this problem, any help, please?

     \cos ^2 ( \frac { \pi }{8} - \frac {1}{2} )
    What is the problem, to evaluate that? Use your claculator, it is very unlikely to have a simple closed form.

    You can mess about with it using trig identities to expand the cos and then simplifying, so:

    \cos^2\left(\frac{\pi/4-1}{2}\right)=\frac{1+\cos(\pi/4-1)}{2}

    Now expand the \cos on the right ..

    CB
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    Umm... I'm still having some difficulties in expanding the right cos...

    I have  \cos ( \frac { \pi } {4} - 1 ) = \frac { \sqrt {2} }{2} cos (1) + \frac { \sqrt {2} }{2} sin (1)

    So I should have  \frac { \sqrt {2} }{4} \cos (1) + \frac { \sqrt {2} }{4}  \sin (1) + \frac {1}{2}

    but then how do I simplify that? The answer should have been  \frac { \pi }{4} according to the book...
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    Umm... I'm still having some difficulties in expanding the right cos...

    I have  \cos ( \frac { \pi } {4} - 1 ) = \frac { \sqrt {2} }{2} cos (1) + \frac { \sqrt {2} }{2} sin (1)

    So I should have  \frac { \sqrt {2} }{4} \cos (1) + \frac { \sqrt {2} }{4}  \sin (1) + \frac {1}{2}

    but then how do I simplify that? The answer should have been  \frac { \pi }{4} according to the book...
    In that case either you have given us the wrong question or the book is wrong

    (\cos(\frac{\pi}{8}-\frac{1}{2}))^2 \approx 0.9885

    while \pi/4 \approx 0.7854

    CB
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