Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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, $$\dfrac{2(\sin a\cdot\cos b+\cos a\cdot\sin b)}{2\cdot\sin b\cdot \cos b}=\dfrac{2\sin (a+b)}{\sin 2b}$$

Hope this helps.

share|cite|improve this answer
Will the downvoter explain? – user21436 Mar 21 '12 at 12:32

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 $$\frac{2(\sin a\cos b+\cos a\sin b)}{2\sin b\cos b}=\frac{2\sin(a+b)}{\sin2b}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.