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Math Help - Trigonometry Identity

  1. #1
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    Trigonometry Identity

    I need help for my homework, thanks.

    1) \tan(X+Y) + \tan(X-Y) = \frac{2\sin(2X)}{\cos(2X) + \cos(2Y)}

    2) If A + B + C = 180 degrees, prove that  \cos(A) + \cos(B) + \cos(C) = 1 + 4\sin{\frac{A}{2}}\sin{\frac{b}{2}}\sin{\frac{c}{2  }}

    Thank you.
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  2. #2
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    Lexington, MA (USA)
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    Hello, elitewarr!

    For the first one, we need three identities:

    . . \tan\theta \:=\:\frac{\sin\theta}{\cos\theta}

    . . \sin A\cos B - \sin B\cos A \:=\:\sin(A - B)

    . . \cos A\cos B \:=\:\tfrac{1}{2}\bigg[\cos(A+B) + \cos(A-B)\bigg]



    1)\;\;\tan(x+y) + \tan(x-y) \:= \:\frac{2\sin2x}{\cos2x + \cos2y}
    Start with the left side:


    \tan(x+y) + \tan(x-y) \:=\:\frac{\sin(x+y)}{\cos(x+y)} + \frac{\sin(x-y)}{\cos(x-y)}


    . . . . . = \;\frac{\sin(x+y)\cos(x-y) + \sin(x-y)\cos(x+y)}{\cos(x+y)\cos(x-y)}


    . . . . . = \;\frac{\sin\bigg[(x+y) + (x-y)\bigg]}{\frac{1}{2}\bigg[\cos[(x+y)+ (x-y)] + \cos[(x+y)-(x-y)]\bigg]}


    . . . . . = \;\frac{\sin2x}{\frac{1}{2}(\cos2x + \cos 2y)}


    . . . . . = \;\frac{2\sin2x}{\cos2x+\cos2y}

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  3. #3
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    OHH!! Never did I ever thought of using sin(a-b).. Haha... Thanks
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