# prove$\frac{ \sin a\vphantom{(}}{\sin b} +\frac{\cos a\vphantom{(}}{\cos b} = \frac{2\sin (a+b)}{\sin 2b}$

I've got this far but don't understand where the $2$ on the numerator comes from: $$\dfrac{\sin a \cos b + \cos a \sin b}{\sin b \cos b}\overset{?}{=}\dfrac{\sin(a+b)}{\sin 2b}$$

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The 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. – Srivatsan Jan 15 '12 at 23:30
so do you times everything by 2? I don't really get it – ejwm Jan 15 '12 at 23:31

$$\dfrac {\sin a}{\sin b}+\dfrac{\cos a}{\cos b}=\dfrac{\sin a\cdot\cos b+\cos a\cdot\sin b}{\sin b\cdot \cos b}$$
After getting this far, you need to observe that, $\boxed{\sin 2b=2 \cdot\sin b \cdot \sin b}$, you'll have to multiply the numerator and denominator by $2$, you'll have the following, $$\dfrac{2(\sin a\cdot\cos b+\cos a\cdot\sin b)}{2\cdot\sin b\cdot \cos b}=\dfrac{2\sin (a+b)}{\sin 2b}$$
So far, you said you've got $$\frac{\sin a\cos b+\cos a\sin b}{\sin b\cos b}.$$ Since $2\sin b\cos b=\sin 2b$, Multiply by $\frac{2}{2}$ to get $$\frac{2(\sin a\cos b+\cos a\sin b)}{2\sin b\cos b}=\frac{2\sin(a+b)}{\sin2b}.$$