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Math Help - Prove(10)

  1. #1
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    Prove(10)

    Prove that sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr  ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}

    attempt

    2sin\frac{3\theta}{2}cos\frac{-\theta}{2}+2sin\frac{7\theta}{2}cos\frac{-\theta}{2}
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  2. #2
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    Quote Originally Posted by Punch View Post
    Prove that sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr  ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}

    attempt

    2sin\frac{3\theta}{2}cos\frac{-\theta}{2}+2sin\frac{7\theta}{2}cos\frac{-\theta}{2}
    Dear Punch,

    sin\theta+sin2\theta+sin3\theta+sin4\theta

    (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)

    2sin\frac{5\theta}{2}cos\frac{3\theta}{2}+2sin\fra  c{5\theta}{2}cos\frac{\theta}{2}

    2sin\frac{5\theta}{2}\left[cos\frac{3\theta}{2}+cos\frac{\theta}{2}\right]

    Can you do it from here?
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear Punch,

    sin\theta+sin2\theta+sin3\theta+sin4\theta

    (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)

    2sin\frac{5\theta}{2}cos\frac{3\theta}{2}+2sin\fra  c{5\theta}{2}cos\frac{\theta}{2}

    2sin\frac{5\theta}{2}\left[cos\frac{3\theta}{2}+cos\frac{\theta}{2}\right]

    Can you do it from here?
    If I am right, solving it will require another use of the factor formula on them.

    However, I am a little puzzled at deciding which terms to group together for use of the factor formula in step one.
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  4. #4
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    Quote Originally Posted by Punch View Post
    If I am right, solving it will require another use of the factor formula on them.

    However, I am a little puzzled at deciding which terms to group together for use of the factor formula in step one.
    Dear Punch,

    Yes, you are correct. Using the factor formula again would give you the answer. However if you have any problems regarding this question please don't hesitate to reply.
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  5. #5
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    Yes, I do have a question. From sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr  ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}, how did you know you had to group it as (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)?
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  6. #6
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    Quote Originally Posted by Punch View Post
    Yes, I do have a question. From sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr  ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}, how did you know you had to group it as (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)?
    Dear Punch,

    As you could see the answer contains a 5\theta. So to get 5\theta you have to group it like that. Suppose if you group (sin\theta+sin3\theta) and (sin2\theta+sin4\theta) then you can't get a 5\theta term.

    Is this answer cleared your problem?
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