Question

\[ \text{If } \sin x+\cos x=a, \]

\[ \text{then } \sin^6x+\cos^6x= \]

Solution

Squaring,

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

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

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

Now,

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

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

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

\[ =1-3\left(\frac{a^2-1}{2}\right)^2 \]

\[ =1-\frac34(a^2-1)^2 \]

Answer

\[ \boxed{1-\frac34(a^2-1)^2} \]