8.1 Q-12

Exercise 8.1 — Question 12

Problem Statement

Prove that: $\sin (\alpha +\beta )\cdot \sin (\alpha -\beta )={\sin }^2\alpha -{\sin }^2\beta$

Solution

Starting with the left-hand side (L.H.S.):

$\text{L.H.S.}=\sin (\alpha +\beta )\cdot \sin (\alpha -\beta )$

Apply the sine addition and subtraction formulas: $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ $\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta$

Multiplying: $=(\sin \alpha \cos \beta +\cos \alpha \sin \beta )(\sin \alpha \cos \beta -\cos \alpha \sin \beta )$

This is of the form $(A+B)(A-B)=A^2-B^2$, where $A=\sin \alpha \cos \beta$ and $B=\cos \alpha \sin \beta$:

$={\sin }^2\alpha {\cos }^2\beta -{\cos }^2\alpha {\sin }^2\beta$

Replace ${\cos }^2\beta =1-{\sin }^2\beta$ and ${\cos }^2\alpha =1-{\sin }^2\alpha$:

$={\sin }^2\alpha (1-{\sin }^2\beta )-(1-{\sin }^2\alpha ){\sin }^2\beta$

$={\sin }^2\alpha -{\sin }^2\alpha {\sin }^2\beta -{\sin }^2\beta +{\sin }^2\alpha {\sin }^2\beta$

$={\sin }^2\alpha -{\sin }^2\beta$

$=\text{R.H.S.}■$

Key Identities Used