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

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(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 = $

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