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Math Help - Another identity proof.

  1. #1
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    Another identity proof.

    One more, guys.

    sin\hspace{5pt}2t - cot\hspace{5pt} t = - cot\hspace{5pt}t\hspace{5pt}cos\hspace{5pt}2t

    On this one, I made a bit of progress, I think.

    Starting from the RHS:

     - \frac{cos t}{sin t} * (2 cos^2 t - 1)

    I think that might be the right direction to start in, but I can't see where it goes from here.

    Thanks again!
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  2. #2
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    Quote Originally Posted by JennyFlowers View Post
    One more, guys.

    sin\hspace{5pt}2t - cot\hspace{5pt} t = - cot\hspace{5pt}t\hspace{5pt}cos\hspace{5pt}2t

    On this one, I made a bit of progress, I think.

    Starting from the RHS:

     - \frac{cos t}{sin t} * (2 cos^2 t - 1)

    I think that might be the right direction to start in, but I can't see where it goes from here.

    Thanks again!
    As you have \sin(2t) on the LHS you are looking to have 2\sin(x)\cos(x) in the numerator. Therefore it is easier if you use the equivalent expression for cos(2x) in terms of sin which is \cos(2x) = 1-2\sin^2(x) instead.

    Spoiler:
    -\frac{\cos(t)}{\sin(t)} (1-2\sin^2(t)) = -\frac{\cos(t)}{\sin(t)} + \frac{2\sin^2(t)\cos(t)}{\sin(t)}

    That all cancels down to 2\sin(t)\cos(t) - \cot(t) = \sin(2t) - \cot(t)
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  3. #3
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    Thank you, that did the trick!
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