If sin alpha + cos alpha = m,then sin^6 alpha | vedclass

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

Question

(B) Given $\sin \alpha + \cos \alpha = m$. Squaring both sides,we get $(\sin \alpha + \cos \alpha)^2 = m^2$. $1 + 2 \sin \alpha \cos \alpha = m^2 \Rig

Vedclass · Accepted Answer

$\frac{4-3(m^2-1)^2}{4}$

Vedclass · Answer

(B) Given $\sin \alpha + \cos \alpha = m$. Squaring both sides,we get $(\sin \alpha + \cos \alpha)^2 = m^2$. $1 + 2 \sin \alpha \cos \alpha = m^2 \Rightarrow 2 \sin \alpha \cos \alpha = m^2 - 1$. Now,$\sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha)^3 + (\cos^2 \alpha)^3$. Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have: $= (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha)$. Since $\sin^2 \alpha + \cos^2 \alpha = 1$,this simplifies to: $= (\sin^2 \alpha + \cos^2 \alpha)^2 - 3 \sin^2 \alpha \cos^2 \alpha$. $= 1 - 3(\sin \alpha \cos \alpha)^2$. $= 1 - 3(\frac{m^2-1}{2})^2$. $= 1 - \frac{3(m^2-1)^2}{4} = \frac{4 - 3(m^2-1)^2}{4}$.

If $\sin \alpha + \cos \alpha = m$,then $\sin^6 \alpha + \cos^6 \alpha = $

Similar Questions

If $\cos \theta + \cos 7\theta + \cos 3\theta + \cos 5\theta = 0$,then $\theta$ is equal to:

If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$,then a possible value of $\sec x$ is

If $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$,then $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta)$ is equal to

If $x = a \cos^3 \theta$ and $y = b \sin^3 \theta$,then:

If $\sin \theta + \cos \theta = 1$,then $\sin \theta \cos \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module