Question

Prove that:
(i) $co{s}^2{45}^{\circ }-si{n}^2{15}^{\circ }=\frac{\sqrt{3}}{4}$
(ii) $Si{n}^2(n+1)A-si{n}^{2}nA=sin(2n+1)AsinA$

Answer 1

(i) $co{s}^2{45}^{\circ }-si{n}^2{15}^{\circ }=\frac{\sqrt{3}}{4}$
${\left(\frac{1}{\sqrt{2}}\right)}^2-si{n}^2{15}^{\circ }$
$=\frac{1}{2}-\left(\frac{1-cos2\times {15}^{\circ }}{2}\right)$
$=\frac{1}{2}-\left(\frac{1-cos{30}^{\circ }}{2}\right)$
$=\frac{1-1+cos{30}^{\circ }}{2}$
$=\frac{cos{30}^{\circ }}{2}$
$=\frac{\sqrt{3}}{1}\times \frac{1}{2}=\frac{\sqrt{3}}{4}$
=RHS
$\therefore LHS=RHS$
Hence proved.
(ii) LHS: $si{n}^2(n+1)A-si{n}^{2}nA$
=sin[(n+1)A+nA]sin[(n+1)A-nA]
=sin[nA+A+nA]sin[nA+A-nA]
=sin(2nA+A)sin(A)
=sin(2n+1)AsinA
=RHS
$\because LHS=RHS$
Hence proved.