I've got this far but don't understand where the $2$ on the numerator comes from: $$\dfrac{\sin a \cos b + \cos a \sin b}{\sin b \cos b}\overset{?}{=}\dfrac{\sin(a+b)}{\sin 2b}$$
prove$\frac{ \sin a\vphantom{(}}{\sin b} +\frac{\cos a\vphantom{(}}{\cos b} = \frac{2\sin (a+b)}{\sin 2b}$
2
$\begingroup$
trigonometry
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2The double-angle identity is $\sin 2b = 2\sin b \cos b$ (not $\sin b \cos b$). So once you get $\frac{\sin a \cos b + \cos a \sin b}{\sin b \cos b} = \frac{\sin (a+b)}{\sin b \cos b}$, multiply the numerator and denominator by $2$ and use the double-angle identity I just mentioned. – 2012-01-15
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0so do you times everything by 2? I don't really get it – 2012-01-15