Prove Trig Identity

Question

For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$

This is what I've tried:

$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\right)}^y2\cos\left(\frac{B-C}2\right)\\ \sin\left(\frac{A-C}2+\frac{A-B}2\right)+\sin\left(\frac{A-C}2-\frac{A-B}2\right)\\ \left(\frac{B-C}2\right)\left[\sin\left(\frac{2A-B-C}2\right)+\sin\left(\frac{B-C}2\right)\right]\\ \cos\left(\frac{B-C}2\right)\sin\left(\frac{2A-B-C}2\right)+\cos\left(\frac{B-C}2\right)\sin\left(\frac{B-C}2\right)$$

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.

Related sum of three sines: https://math.stackexchange.com/q/4518662 — Davius, Aug 27 '22 at 22:00

score 2 · Accepted Answer · answered Dec 18 '13 at 04:37

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. $$ 2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{2A - B - C}{2}\right) + 2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{B-C}{2}\right) \\ = \sin\left(\frac{B-C}{2} + \frac{2A - B - C}{2}\right) - \sin\left(\frac{B-C}{2} - \frac{2A - B - C}{2}\right) + \sin(B - C) \\ = \sin(A - C) - \sin(-A + B) + \sin(B - C) \\ = \sin(A - C) + \sin(A - B) + \sin(B - C) $$

score 0 · Answer 2 · answered Dec 18 '13 at 04:26

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

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

answered Dec 18 '13 at 04:26

lab bhattacharjee

274,582

I don't see how I could use the formulas on sin(2A -B-C /2) + sin (B - C / 2) – Dec 18 '13 at 04:41
Nvm,figured it out. Thanks anyway. – Dec 18 '13 at 04:49
@Pallas, nice to hear that. But how? – lab bhattacharjee Dec 18 '13 at 13:26

score 0 · Answer 3 · answered Dec 18 '13 at 06:08

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

Prove Trig Identity

3 Answers3