If \[ \sin x+\cos x=m \] Prove that \[ \sin^6 x+\cos^6 x = \frac{4-3(m^2-1)^2}{4} \] where \[ m^2\le 2 \]

Solution:

\[ \sin^6 x+\cos^6 x \]

\[ = (\sin^2 x)^3+(\cos^2 x)^3 \]

\[ = (\sin^2 x+\cos^2 x) (\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x) \]

\[ = \sin^4 x+\cos^4 x-\sin^2 x\cos^2 x \]

\[ = (\sin^2 x+\cos^2 x)^2 -3\sin^2 x\cos^2 x \]

\[ = 1-3\sin^2 x\cos^2 x \]

Now,

\[ \sin x+\cos x=m \]

Squaring,

\[ m^2 = \sin^2 x+\cos^2 x+2\sin x\cos x \]

\[ = 1+2\sin x\cos x \]

\[ \sin x\cos x = \frac{m^2-1}{2} \]

Therefore,

\[ \sin^2 x\cos^2 x = \left(\frac{m^2-1}{2}\right)^2 \]

Substituting,

\[ \sin^6 x+\cos^6 x = 1- 3\left(\frac{m^2-1}{2}\right)^2 \]

\[ = \frac{4-3(m^2-1)^2}{4} \]

Hence proved.