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Math Help - Hard Trigonometry

  1. #1
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    Hard Trigonometry

    Evaluate the sum \sin^2 1^{\circ}+\sin^2 2^{\circ}+\sin^2 3^{\circ}+\ldots+\sin^2 90^{\circ}.

    Note: The numbers are in degrees, example: \sin 90^{\circ}=1
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  2. #2
    Moo
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    Hello,

    Let N be this sum.

    If you replace the sinus with cosine, you'll get something interesting :

    \sin^2(x)=1-\cos^2(x)

    So replace for each term.

    You'll have :

    N=(1-\cos^2 1) + (1-\cos^2 2) + \dots + (1-\cos^2 90)  = 90-(\cos^2 1 + \cos^2 2 + \dots + \cos^2 90)

    But we know that \cos(x)=\sin(90-x) \Longrightarrow \cos^2 (x)=\sin^2(90-x)

    So if we replace again :

    N=90-(\sin^2 89 + \sin^2 88 + \dots + \sin^2 0) (1)

    The aim will be to retrieve N in the brackets. So tihere is sin^2 90 missing and an extra  \sin^2 0.

    sin(0)=0, so we can remove  \sin^2 0. As sin(90)=1, sin^2 90=1

    By using (1) :

    N=90-((\sin^2 90 + \sin^2 89 + \sin^2 88 + \dots + \sin^2 1) - \sin^2 90)

    N=90-(N-1)

    And then you have N
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