Question

If $(\cos \theta)^{6}+(\sin \theta)^{6} \equiv a_{0}+a_{1} \cos 4 \theta$ and $a_{0} a_{1}=\frac{m}{n}$ (where $\left.\frac{m}{n} \in Q\right)$ is in its lowest form, then $(m+n)$ equals

Answer 1

$\left(\cos ^{2} \theta+\sin ^{2} \theta\right)^{3}-3 \sin ^{2} \theta \cos ^{2} \theta$
$1-\frac{3}{4} \sin ^{2} 2 \theta=1-\frac{3}{8}[1-\cos 4 \theta]=\frac{5}{8}+\frac{3}{8} \cos 4 \theta$
$\Rightarrow a_{0} a_{1}=\frac{5}{8} \times \frac{3}{8}=\frac{15}{64}=\frac{m}{n}$
$\Rightarrow(m +n)=79$