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Math Help - Trigonometry Triangle Problem

  1. #1
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    Trigonometry Triangle Problem

    Hi it's me again..

    I'm having a bit of trouble visualizing this question. It's the last one in my "Holiday Homework package, and I have no idea what it means...


    In any triangle ABC, prove that:

    (a-b)/b= (2 sin(C/2) sin((A-B)/2))/ sinB


    P.S: I didn't want to revive a dead thread, but I would like to thank Soroban for answering the question on my last thread!
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  2. #2
    MHF Contributor red_dog's Avatar
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    From the sine rule \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R

    we have a=2R\sin A, \ b=2R\sin B

    Then \frac{a-b}{b}=\frac{\sin A-\sin B}{\sin B}=

    =\frac{2\sin\frac{A-B}{2}\cos\frac{A+B}{2}}{\sin B}=\frac{2\sin\frac{A-B}{2}\sin\frac{C}{2}}{\sin B}
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  3. #3
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    THank you very much for posting an answer. I understand the first part of the question very clearly. I just have a question about the last part:

    How does cos((A+B/2) become sin (C/2)?

    Thanks!
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by KelvinScale View Post
    THank you very much for posting an answer. I understand the first part of the question very clearly. I just have a question about the last part:

    How does cos((A+B/2) become sin (C/2)?

    Thanks!
    \cos\frac{A+B}{2}=\cos\frac{\pi-C}{2}=\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)=\sin\frac{C}{2}

    because \cos\left(\frac{\pi}{2}-x\right)=\sin x
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