sin ^{6} theta+cos ^{6} theta is equal to | vedclass

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

Question

(C) We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 	heta = (\sin^2

Vedclass · Accepted Answer

$1-3 \sin ^{2} 	heta \cos ^{2} 	heta$

Vedclass · Answer

(C) We know that $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3$.Using the identity,we get:$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta + \cos^2 	heta)^3 - 3(\sin^2 	heta)(\cos^2 	heta)(\sin^2 	heta + \cos^2 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we substitute this value:$\sin^6 	heta + \cos^6 	heta = (1)^3 - 3 \sin^2 	heta \cos^2 	heta (1)$.Therefore,$\sin^6 	heta + \cos^6 	heta = 1 - 3 \sin^2 	heta \cos^2 	heta$.

$\sin ^{6} \theta+\cos ^{6} \theta$ is equal to

Similar Questions

If $\tan^2 \theta = 2\tan^2 \phi + 1$,then $\cos 2\theta + \sin^2 \phi$ equals

If $\sin \theta = \frac{24}{25}$ and $\theta$ lies in the second quadrant,then $\sec \theta + \tan \theta = $

$\tan 15^\circ = $

In a right-angled triangle,the hypotenuse is $2 \sqrt{2}$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

Let ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ for $k = 1, 2, 3, ...$. Then for all $x \in R$,the value of $f_4(x) - f_6(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module