Question

How do you prove $\sin (\mathrm{A}+\mathrm{B})\times \sin (\mathrm{A}-\mathrm{B})={\sin }^2\left(\mathrm{A}\right)-{\sin }^2\left(\mathrm{B}\right)$?

Answer 1

Prove the given equation

We know that $\sin (\mathrm{A}+\mathrm{B})=\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)+\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)$

And, $\sin (\mathrm{A}-\mathrm{B})=\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)-\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)$

$\therefore$$\sin (\mathrm{A}+\mathrm{B})\times \sin (\mathrm{A}-\mathrm{B})=$$\left[\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)+\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)\right]$$\left[\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)-\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)\right]$

$={\left[\sin \left(\mathrm{A}\right)\cos \left(\mathrm{B}\right)\right]}^2-{\left[\cos \left(\mathrm{A}\right)\sin \left(\mathrm{B}\right)\right]}^2$ $\left[\because \left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2\right]$

$=\left[{\sin }^2\left(\mathrm{A}\right){\cos }^2\left(\mathrm{B}\right)\right]-\left[{\cos }^2\left(\mathrm{A}\right){\sin }^2\left(\mathrm{B}\right)\right]$

$={\sin }^2\left(\mathrm{A}\right)\left[1-{\sin }^2\left(\mathrm{B}\right)\right]-\left[1-{\sin }^2\left(\mathrm{A}\right)\right]{\sin }^2\left(\mathrm{B}\right)$ $\left[\because {\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)=1\Rightarrow 1-{\sin }^2\left(\mathrm{\theta }\right)={\cos }^2\left(\mathrm{\theta }\right)\right]$

$={\sin }^2\left(\mathrm{A}\right)-\left[{\sin }^2\left(\mathrm{A}\right)\times {\sin }^2\left(\mathrm{B}\right)\right]-{\sin }^2\left(\mathrm{B}\right)+\left[{\sin }^2\left(\mathrm{A}\right)\times {\sin }^2\left(\mathrm{B}\right)\right]$

$\Rightarrow$$\sin (\mathrm{A}+\mathrm{B})\times \sin (\mathrm{A}-\mathrm{B})=$${\sin }^2\left(\mathrm{A}\right)-{\sin }^2\left(\mathrm{B}\right)$

Hence proved