Question:
Prove that: $\cos ^{6} A-\sin ^{6} A=\cos 2 A\left(1-\frac{1}{4} \sin ^{2} 2 A\right)$
Solution:
LHS $=\cos ^{6} A-\sin ^{6} A$
$=\left(\cos ^{2} A\right)^{3}-\left(\sin ^{2} A\right)^{3}$
$=\left(\cos ^{2} A-\sin ^{2} A\right)\left(\cos ^{4} A+\sin ^{2} A \cdot \cos ^{2} A+\sin ^{4} A\right)$
$=\cos 2 A\left(\cos ^{4} A+2 \sin ^{2} A \cos ^{2} A+\sin ^{4} A-\sin ^{2} A \cos ^{2} A\right)$
$=\cos 2 A\left\{\left(\sin ^{2} A+\cos ^{2} A\right)^{2}-\frac{1}{4} \times 4 \sin ^{2} A \cos ^{2} A\right\}$
$=\cos 2 A\left\{\left(\sin ^{2} A+\cos ^{2} A\right)^{2}-\frac{1}{4}(2 \sin A \cos A)^{2}\right\}$
$=\cos 2 A\left\{1-\frac{1}{4}(\sin 2 A)^{2}\right\}$
$=\cos 2 A\left\{1-\frac{1}{4} \sin ^{2} 2 A\right\}=\mathrm{RHS}$
Hence proved.