Question

If $\sin x+\cos x=a$, then find the value of ${\sin }^{6}x+{\cos }^{6}x$.

Answer 1

Given: $\sin x+\cos x=a$

Squaring on both sides, we get

$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=a^2 \\ \Rightarrow 1+2\sin x\cos x=a^2 \\ \Rightarrow \sin x\cos x=\frac{a^2-1}{2}.....\left(1\right) \end{gathered}$$

Now,

$$\begin{gathered} {\sin }^{6}x+{\cos }^{6}x \\ ={\left({\sin }^{2}x+{\cos }^{2}x\right)}^3-3{\sin }^{2}x{\cos }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right) \\ =1-3{\left(\frac{a^2-1}{2}\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{4-3{\left(a^2-1\right)}^2}{4} \end{gathered}$$

Hence, the required value is $\frac{1}{4}\left[4-3{\left(a^2-1\right)}^2\right]$.