sin^6 theta + cos^6 theta + 3 sin^2 theta cos | vedclass

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

Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.Let $a = \sin^2 	heta$ and $b = \cos^2 	heta$.Then,$\sin^6 	heta + \cos^6 	heta = (\sin^2 	heta)^3 + (\cos^2 	heta)^3 = (\sin^2 	heta + \cos^2 	heta)^3 - 3 \sin^2 	heta \cos^2 	heta (\sin^2 	heta + \cos^2 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin^6 	heta + \cos^6 	heta = (1)^3 - 3 \sin^2 	heta \cos^2 	heta (1) = 1 - 3 \sin^2 	heta \cos^2 	heta$.Substituting this into the original expression:$\sin^6 	heta + \cos^6 	heta + 3 \sin^2 	heta \cos^2 	heta = (1 - 3 \sin^2 	heta \cos^2 	heta) + 3 \sin^2 	heta \cos^2 	heta = 1$.Alternatively,by putting $	heta = 0^\circ$,we get $\sin^6(0^\circ) + \cos^6(0^\circ) + 3 \sin^2(0^\circ) \cos^2(0^\circ) = 0 + 1 + 0 = 1$.

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

Similar Questions

If $\sin 2\theta + \sin 2\phi = \frac{1}{2}$ and $\cos 2\theta + \cos 2\phi = \frac{3}{2}$,then $\cos^2(\theta - \phi) =$

If $\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_3) + \cos(\theta_4) = -4$,then the value of $\cot(\frac{\theta_1}{2}) + \cot(\frac{\theta_2}{2}) + \cot(\frac{\theta_3}{2}) + \cot(\frac{\theta_4}{2}) = $

Let $f(\theta) = \sin \theta (\sin \theta + \sin 3\theta)$,then $f(\theta)$

The number of all possible integral values of $n > 2$ such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$ is:

If $a \cos 2\theta + b \sin 2\theta = c$ has $\alpha$ and $\beta$ as its solutions,then the value of $\tan \alpha + \tan \beta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module