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Math Help - trig identity

  1. #1
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    Exclamation trig identity

    Prove that:

    cos(2x-3y)+ cos3y = cot a
    sin(2x-3y)+ sin3y
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  2. #2
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    Hello, math123456!

    You have a typo . . .


    Prove that: . \frac{\cos(2x-3y)+ \cos3y}{\sin(2x-3y) + \sin3y} \:=\:\cot{\color{red}x}
    We need two Sum-to-Product identities:

    . . \sin A + \sin B \:=<br />
\:2\sin\left(\tfrac{A+B}{2}\right)\cos\left(\tfrac  {A-B}{2}\right)

    . . \cos A + \cos B \:=\:2\cos\left(\tfrac{A+B}{2}\right)\cos\left(\tf  rac{A-B}{2}\right)



    Numerator:
    \cos(2x-3y)+\cos3y \:=\:2\cos\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\cos x\cos(x-3y)


    Denominator:
    \sin(2x-3y)+\sin3y \:=\:2\sin\left(\frac{2x-3y+3y}{2}\right)\cos\left(\frac{2x-3y-3y}{2}\right) \:=\:2\sin x\cos(x - 3y)


    The fraction becomes: . \frac{2\,\cos x\,\cos(x - 3y)}{2\,\sin x\,\cos(x - 3y)} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x

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  3. #3
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    Quote Originally Posted by math123456 View Post
    Prove that:

    cos(2x-3y)+ cos3y = cot a
    sin(2x-3y)+ sin3y
    Use the identities

    \sin{a} + \sin{b} = 2\sin{\frac{a+b}{2}}\cos{\frac{a-b}{2}}

    and

    \cos{a} + \cos{b} = 2\cos{\frac{a+b}{2}}\cos{\frac{a-b}{2}}.


    So \frac{\cos{2x - 3y} + \cos{3y}}{\sin{2x - 3y} + \sin{3y}}

     = \frac{2\cos{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}{2\sin{\frac{2x - 3y + 3y}{2}}\cos{\frac{2x - 3y - 3y}{2}}}

     = \frac{\cos{x}\cos{(x - 3y)}}{\sin{x}\cos{(x - 3y)}}

     = \frac{\cos{x}}{\sin{x}}

     = \cot{x}.
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