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

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

Question

(B) We are given the equation: ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$. Using the identity ${a^3} + {b^3} = {(a + b)^3} - 3ab(a +

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(B) We are given the equation: ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$. Using the identity ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$,let $a = {\cos ^2}\alpha$ and $b = {\sin ^2}\alpha$. Then,${{\cos ^6}\alpha + {\sin ^6}\alpha = {(\cos ^2}\alpha + {\sin ^2}\alpha)^3 - 3{\cos ^2}\alpha {\sin ^2}\alpha (\cos ^2}\alpha + {\sin ^2}\alpha)$. Since ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$,this simplifies to $1 - 3{\cos ^2}\alpha {\sin ^2}\alpha$. We know that $2\sin \alpha \cos \alpha = \sin 2\alpha$,so ${\sin ^2}2\alpha = 4{\sin ^2}\alpha {\cos ^2}\alpha$,which implies ${\sin ^2}\alpha {\cos ^2}\alpha = \frac{1}{4}{\sin ^2}2\alpha$. Substituting this back into the equation: $1 - 3(\frac{1}{4}{\sin ^2}2\alpha) + K{\sin ^2}2\alpha = 1$. $K{\sin ^2}2\alpha - \frac{3}{4}{\sin ^2}2\alpha = 0$. $(K - \frac{3}{4}){\sin ^2}2\alpha = 0$. For this to hold for all $\alpha$,we must have $K = \frac{3}{4}$.

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

Similar Questions

If $x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ and $x \sin \alpha - y \cos \alpha = 0$,then $x^2 + y^2 = $

The value of $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$ is

The number of real solutions of the equation $(\sin x - x)(\cos x - x^2) = 0$ is

If $a \sin^2 \theta + b \cos^2 \theta = c$,then $\tan^2 \theta$ is equal to

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0$,then find the value of $5 \sin 2A + 3 \sin A + 4 \cos A$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module