8.1 Q-13

Exercise 8.1 — Question 13

Problem Statement

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

Solution

We use the fundamental law of trigonometry (sine addition and subtraction formulas):

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

Left-Hand Side (L.H.S.):

$\sin (\alpha +\beta )\cdot \sin (\alpha -\beta )$

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

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

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

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

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

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

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

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

= R.H.S. $■$

Key Identity Used

The difference of squares pattern combined with the Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$ is the core technique here.