If {cos ^6}alpha + {sin ^6}alpha + K,{sin ^2} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) Since ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$

using formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ and on solving,

w

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(b) Since ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$

using formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ and on solving,

we get the required result $i.e.$ $K = \frac{3}{4}$.

If ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ then $K =$

Similar Questions

If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to

$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $

If $A$ and $B$ are complimentary angles, then :

The value of the expression $(sinx + cosecx)^2 + (cosx + secx)^2 - ( tanx + cotx)^2$ wherever defined is equal to

The value of $tan^{-1} (\frac{sin2 -1}{cos2})$ is equal to:-