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 $\frac{m}{n}\in Q)$ is in its lowest form, then $(m+n)$ equals

Answer 1

Answer: 79

${\left({\cos }^2\theta +{\sin }^2\theta \right)}^3-3{\sin }^2\theta {\cos }^2\theta$
$1-\frac{3}{4}{\sin }^{2}2\theta =1-\frac{3}{8}[1-\cos 4\theta ]=\frac{5}{8}+\frac{3}{8}\cos 4\theta$
$\Rightarrow a_0{a}_1=\frac{5}{8}\times \frac{3}{8}=\frac{15}{64}=\frac{m}{n}$
$\Rightarrow (m+n)=79$