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Math Help - Trigonometric Identities problem

  1. #1
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    Question Trigonometric Identities problem

    I would be very grateful if someone could solve this question for me???

    Using the identities for sin (A+B) and cos^2(theta)+sin^2(theta)=1

    Find sine 65 degrees?
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  2. #2
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    Lightbulb Failed attempt

    okay we get started by listing what may be of help here:
    \sin60^\circ=\frac{\sqrt3}2;
    \sin30^\circ=\frac12;
    \sin45^\circ=\frac{\sqrt2}2;
    with the trigonometric identities:
    \sin(a\pm b)=\sin a\cos b\pm\sin b\cos a
    \sin^2 a+\cos^2 a = 1

    First we assume
    y=\sin20^\circ;;
    therefore we have
    \cos20^\circ=\sqrt{1-y^2},
    together with the application of the identities introduced above, which results:
    \sin40^\circ=2\sin20^\circ\cos20^\circ=2y\sqrt{1-y^2},
    \cos40^\circ=\sqrt{1-\sin^240^\circ}=\dots=1-2y^2
    (note that y is less than \sin45^\circ)

    The identities also yield that
    \sin60^\circ=\sin(20^\circ+40^\circ)
    =\sin20^\circ\cos40^\circ+\cos20^\circ\sin40^\circ  =\dots
    =-4y^3+3y=\frac{\sqrt3}2
    Therefore we may find the only real value of y (which isn't easy)
    y\approx 0.6428
    Hence
    \sin65^\circ\sin(45^\circ+20^\circ)=\dots=\frac{\s  qrt2}2(y+\sqrt{1-y^2})=0.9063

    The bad news is that, I haven't yet figured out how to solve the equation without a calulator. So good luck on that, and I'll put my findings here as soon as I'm getting something.
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  3. #3
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    I think there is not really a nice answer to this problem. GAVREED2, are you sure there's no typo?
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  4. #4
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    Quote Originally Posted by helopticor View Post
    I think there is not really a nice answer to this problem. GAVREED2, are you sure there's no typo?
    see this "second" post ...

    http://www.mathhelpforum.com/math-he...le-153405.html
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