Question

If ${\sin }^6\theta +{\cos }^6\theta =1-k{\sin }^{2}2\theta$, then find the value of $k$

Answer 1

We have,

${\sin }^6\theta +{\cos }^6\theta =1-k{\sin }^{2}2\theta$

By using

$a^3+b^3=(a^2+b^2{)}^3-3a^2.b^2(a^2+b^2)$

$=({\sin }^2\theta +{\cos }^2\theta {)}^3-3{\sin }^2\theta .{\cos }^2\theta ({\sin }^2\theta +{\cos }^2\theta )=1-k{\sin }^{2}2\theta$

$1^3-3{\sin }^2\theta .{\cos }^2\theta =1-k{\sin }^{2}2\theta$

$k=\frac{3{\sin }^2\theta .{\cos }^2\theta }{{\sin }^{2}2\theta }$

$=\frac{3{\sin }^2\theta .{\cos }^2\theta }{2{\sin }^2\theta .co{s}^2\theta }$

$=\frac{3}{2}$

Therefore,

$k=\frac{3}{2}$