# Prove Trig Identity

sorry if this question isn't high quality. I don't understand #11 here:http://imgur.com/QQ5cmOb This is what I've tried http://m.imgur.com/3keNf1o

I know it's basically turning sum into product and I previously tried to derive the LHS the same way the normal sum to product identity works but I couldn't figure out how. I tried to use the sum-product identity on RHS but couldn't quite make it work.

Thanks.

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You are missing a factor of $2$ in the last line of your answer. Anyway, you were on the right track. You just need to use the same type of identity (as you used in the beginning) for the first term and the double angle formula for the second. $$2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{2A - B - C}{2}\right) + 2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{B-C}{2}\right) \\ = \sin\left(\frac{B-C}{2} + \frac{2A - B - C}{2}\right) - \sin\left(\frac{B-C}{2} - \frac{2A - B - C}{2}\right) + \sin(B - C) \\ = \sin(A - C) - \sin(-A + B) + \sin(B - C) \\ = \sin(A - C) + \sin(A - B) + \sin(B - C)$$

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Now take out the common factor $\displaystyle\cos\frac{B-C}2$

and apply Prosthaphaeresis Formulas on $\displaystyle\sin\frac{2A-B-C}2+\sin\frac{B-C}2$

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I don't see how I could use the formulas on sin(2A -B-C /2) + sin (B - C / 2) – Pallas Dec 18 '13 at 4:41
Nvm,figured it out. Thanks anyway. – Pallas Dec 18 '13 at 4:49
@Pallas, nice to hear that. But how? – lab bhattacharjee Dec 18 '13 at 13:26

\begin{align} 4\cos(\phi)\cos(\theta)\sin(\phi+\theta) &=2\big[\cos(\phi-\theta)+\cos(\phi+\theta)\big]\sin(\phi+\theta)\tag{1}\\ &=\sin(2\phi)+\sin(2\theta)+\sin(2\phi+2\theta)\tag{2} \end{align} Justification:
$(1)$: $2\cos(\phi)\cos(\theta)=\cos(\phi-\theta)+\cos(\phi+\theta)$

$(2)$: $\begin{array}{}2\cos(\phi-\theta)\sin(\phi+\theta)=\sin(2\phi)+\sin(2\theta)\\2\cos(\phi+\theta)\sin(\phi+\theta)=\sin(2\phi+2\theta)\end{array}$

Now set $\phi=\frac{\alpha-\beta}{2}$ and $\theta=\frac{\beta-\gamma}{2}$

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