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Math Help - Trig Lab Proof

  1. #1
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    Exclamation Trig Lab Proof

    Please help me get started or to complete this proof...have been working on six problems all week and have this one left and running out of time!

    [sin (a+b)]/[sin a + sin b] = [sin a -sin b]/[sin (a-b)]

    Any guidance would be appreciated!
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  2. #2
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    Hello aihal210

    Welcome to Math Help Forum!
    Quote Originally Posted by aihal210 View Post
    Please help me get started or to complete this proof...have been working on six problems all week and have this one left and running out of time!

    [sin (a+b)]/[sin a + sin b] = [sin a -sin b]/[sin (a-b)]

    Any guidance would be appreciated!
    Do you know the following formulae:
    \sin2\theta=2\sin \theta \cos \theta

    \sin a + \sin b=2\sin\tfrac12(a+b)\cos\tfrac12(a-b)

    \sin a - \sin b=2\sin\tfrac12(a-b)\cos\tfrac12(a+b)
    These are the ones to use. Start like this:

    \frac{\sin(a+b)}{\sin a + \sin b}=\frac{2\sin\tfrac12(a+b)\cos\tfrac12(a+b)}{2\si  n\tfrac12(a+b)\cos\tfrac12(a-b)}
    =...

    Simplify the fraction. Then multiply top-and-bottom by 2\sin\tfrac12(a-b), and use the first and last of the three formulae above.

    Can you complete it now?


    Grandad
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  3. #3
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    Hi Grandad, Thanks for responding.

    I did know those formulas and have been trying to work with them. I don't understand how you got the bottom on the right side to be that?

    As far as I knew,

    I thought that sin (a-b) could only be changed into sinacosb-sinbcosa using the sum and difference identities...

    Am I missing something?
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  4. #4
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    Hello aihal210
    Quote Originally Posted by aihal210 View Post
    Hi Grandad, Thanks for responding.

    I did know those formulas and have been trying to work with them. I don't understand how you got the bottom on the right side to be that?

    As far as I knew,

    I thought that sin (a-b) could only be changed into sinacosb-sinbcosa using the sum and difference identities...

    Am I missing something?
    Are you sure you are thinking of the correct formula? The one I've used there is sometimes called the 'Sum to Product' formula. You'll find some more information here.

    Don't confuse this with \sin(a-b) = \sin a \cos b - \cos a\sin b. That's a different formula altogether (although it is one of the ones that is used to derive the Sum to Product formulae).

    Grandad
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