Hello aihal210
Welcome to Math Help Forum! Originally Posted by
aihal210 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:
$\displaystyle \sin2\theta=2\sin \theta \cos \theta$
$\displaystyle \sin a + \sin b=2\sin\tfrac12(a+b)\cos\tfrac12(a-b)$
$\displaystyle \sin a - \sin b=2\sin\tfrac12(a-b)\cos\tfrac12(a+b)$
These are the ones to use. Start like this:
$\displaystyle \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)}$$\displaystyle =...$
Simplify the fraction. Then multiply top-and-bottom by $\displaystyle 2\sin\tfrac12(a-b)$, and use the first and last of the three formulae above.
Can you complete it now?
Grandad