Further simplifying a trigonometric expression.

Question

How do I prove that $\dfrac{\sin(\alpha- \beta)}{\sin \alpha - \sin \beta}= \dfrac{\sin\alpha +\sin \beta}{\sin (\alpha + \beta)}$?

Using double angle and addition formulas, I simplified the LHS to $\dfrac{\cos\dfrac{\alpha - \beta}{2}}{\cos\dfrac{\alpha +\beta}{2}}$

I would suggest clearing denominators and proving what you are left with. — Ted Shifrin, Nov 07 '17 at 06:55

score 1 · Accepted Answer · answered Nov 07 '17 at 06:53

1

Method$\#1:$

Use Prosthaphaeresis Formulas and double angle formula: $$\sin2x=2\sin x\cos x$$ in the Right Hand Side as well.

Method$\#2:$

Alternatively using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ , $$\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B=?$$

answered Nov 07 '17 at 06:53

lab bhattacharjee

274,582

I got the simplified answer using Prosthaphaeresis formula. – Archer Nov 07 '17 at 06:58
@Abcd, Nice to hear that.But I think the problem came into being by method $#2$ – lab bhattacharjee Nov 07 '17 at 06:58

score 1 · Answer 2 · answered Nov 07 '17 at 07:09

$\dfrac{\sin(\alpha - \beta )}{\sin\alpha - \sin\beta}= \dfrac{\sin(\alpha - \beta )\sin (\alpha +\beta)}{(\sin\alpha - \sin\beta)(\sin (\alpha +\beta))}$

Recall that $\sin(\alpha - \beta)\sin(\alpha + \beta)= \sin^2 \alpha - \sin^2\beta$.

Simplifying, we get:

$\dfrac{\sin\alpha+\sin\beta}{\sin(\alpha +\beta)} =RHS$

Q.E.D.

Further simplifying a trigonometric expression.

2 Answers2