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Math Help - cos32 + cos58 + cos20 + cos70

  1. #1
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    cos32 + cos58 + cos20 + cos70

    I ran into this problem on a worksheet:

    Simplify and Evaluate: (without using a calculator)
    cos32 + cos58 + cos20 + cos70


    I have tried using difference of sine and difference of cosine by
    cos (30 +2) = cos 30cos2 - sin30sin2
    etc.. with the others

    still nothing comes to mind


    anyone want to give it a shot?
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  2. #2
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    Quote Originally Posted by unit
    I ran into this problem on a worksheet:

    Simplify and Evaluate: (without using a calculator)
    cos32 + cos58 + cos20 + cos70

    I have tried using difference of sine and difference of cosine by
    cos (30 +2) = cos 30cos2 - sin30sin2
    etc.. with the others
    still nothing comes to mind

    anyone want to give it a shot?
    Hello,

    1. use \cos(\alpha)+\cos(\beta)=2\cos \frac{\alpha+\beta}{2}\ \cdot \  2\cos \frac{\alpha-\beta}{2}

    2. The first 2 angles add up to 90, the 2nd two too. So the first factor on the RHS of the equation will be \sqrt{2}

    I hope this will help you a little bit further on. And now I've to hurry: Duty is calling!

    Greetings

    EB
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  3. #3
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    Are there any other ways of doing this problem?

    I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

    Maybe you could show me how to derive this? All help greatly appreciated! Thanks!
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  4. #4
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    Quote Originally Posted by unit
    Are there any other ways of doing this problem?

    I am not familiar with the identity cosA + cos B = 2cos((a+b)/2)*2cos((a-b)/2)

    Maybe you could show me how to derive this? All help greatly appreciated! Thanks!
    Notice that,
    2\cos x\cos y=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y
    Thus, using the formula:
    \cos(x\pm y)=\cos x\cos y\mp\sin x\sin y
    Thus, we have that:
    2\cos x\cos y=\cos(x+y)+\cos(x-y)
    Now let,
    A=x+y
    B=x-y
    Thus,
    x=\frac{A+B}{2}
    y=\frac{A-B}{2}
    Substituting this we get,
    \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
    Q.E.D.
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