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Math Help - Prove my mathematics induction, trigonometry

  1. #1
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    Red face Prove my mathematics induction, trigonometry

    if u can do this, you are a genius

    Prove by mathematical induction:

    \sum_{k=1}^{n}sink\theta  = \frac{sin[\frac{1}{2}(n+1)\theta ]sin(\frac{1}{2}n\theta )}{sin(\frac{1}{2}\theta )}

    For n =1,
    LHS = sin\theta
    RHS =  \frac{sin(\frac{1}{2}(2\theta ))sin(\frac{1}{2})\theta }{sin(\frac{1}{2}\theta )} = sin\theta
    therefore, it is true for n =1

    assume that it is true for n = a,
     \sum_{k=1}^{a}sina\theta  = \frac{sin[\frac{1}{2}(a+1)\theta ]sin(\frac{1}{2}a\theta )}{sin(\frac{1}{2}\theta )}

    for n = a + 1

    Prove:   \sum_{k=1}^{a}sina\theta  = \frac{sin[\frac{1}{2}(a+2)\theta ]sin(\frac{1}{2}(a+1)\theta )}{sin(\frac{1}{2}\theta )}

    I don't know how to do the next step. I'm really bad with trigonometry.

    HELP!

    Thank you in advance
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by differentiate View Post
    if u can do this, you are a genius

    Prove by mathematical induction:

    \sum_{k=1}^{n}sink\theta  = \frac{sin[\frac{1}{2}(n+1)\theta ]sin(\frac{1}{2}n\theta )}{sin(\frac{1}{2}\theta )}

    For n =1,
    LHS = sin\theta
    RHS =  \frac{sin(\frac{1}{2}(2\theta ))sin(\frac{1}{2})\theta }{sin(\frac{1}{2}\theta )} = sin\theta
    therefore, it is true for n =1

    assume that it is true for n = a,
     \sum_{k=1}^{a}sina\theta  = \frac{sin[\frac{1}{2}(a+1)\theta ]sin(\frac{1}{2}a\theta )}{sin(\frac{1}{2}\theta )}

    for n = a + 1

    Prove:   \sum_{k=1}^{a}sina\theta  = \frac{sin[\frac{1}{2}(a+2)\theta ]sin(\frac{1}{2}(a+1)\theta )}{sin(\frac{1}{2}\theta )}

    I don't know how to do the next step. I'm really bad with trigonometry.

    HELP!

    Thank you in advance
    By assumption:

     S_a(\theta)=\sum_{k=1}^{a}\sin(k\theta)  = \frac{\sin[\frac{1}{2}(a+1)\theta ]\sin(\frac{1}{2}a\theta )}{\sin(\frac{1}{2}\theta )}

    Now:

     S_{a+1}(\theta)=\sum_{k=1}^{a+1}\sin(k\theta)=\sum  _{k=1}^{a}\sin(k\theta)+\sin[(a+1)\theta]    = \frac{\sin[\frac{1}{2}(a+1)\theta ]\sin(\frac{1}{2}a\theta )}{\sin(\frac{1}{2}\theta )} +\sin[(a+1)\theta]

    Now you simplify the right most expression.

    CB
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