Question

Prove that $\sin (2n+1)A.\sin A={\sin }^2(n+1)A-{\sin }^{2}nA$

Answer 1

Consider, R.H.S$={\sin }^2(n+1)A-{\sin }^{2}nA$

$=(\sin (n+1)A-\sin nA)(\sin (n+1)A+\sin nA)$

$=\left(2\cos \left(\frac{n+1+n}{2}\right)A\sin \left(\frac{n+1-n}{2}\right)A\right)\left(2\sin \left(\frac{n+1+n}{2}\right)A\cos \left(\frac{n+1-n}{2}\right)A\right)$

$=\left(2\cos \left(\frac{(2n+1)A}{2}\right)\sin \left(\frac{A}{2}\right)\right)\left(2\sin \left(\frac{(2n+1)A}{2}\right)\cos \left(\frac{A}{2}\right)\right)$

$=(2\cos \left(\frac{(2n+1)A}{2}\right)\times \sin \left(\frac{(2n+1)A}{2}\right))\left(2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)\right)$

$=\sin (2n+1)\sin A$ since $2\sin A\cos A=\sin 2A$

$=$L.H.S

Hence proved.