Question

If sin x + cos x = a, then sin⁶x + cos⁶x = __________.

Answer 1

Given sin x + cos x = a
Squaring both sides, (sin x + cos x)² = a²
$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=a^2 \\ \mathrm{i}.\mathrm{e}1+2\sin x\cos x=a^2 \\ \mathrm{i}.\mathrm{e}2\sin x\cos x=a^2-1\left(1\right) \end{gathered}$$
Using identity, we have
$$\begin{gathered} a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right) \\ {\sin }^{6}x+{\cos }^{6}x={\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3 \\ ={\left({\sin }^{2}x+{\cos }^{2}x\right)}^3-3{\cos }^{2}x{\sin }^{2}x\left({\sin }^{2}x+{\cos }^{2}x\right) \\ ={\left(1\right)}^3-3{\cos }^{2}x{\sin }^{2}x\left(1\right) \\ =1-3{\left(\cos x\sin x\right)}^2 \\ =1-3{\left(\frac{a^2-1}{2}\right)}^2\left(\mathrm{from}\left(1\right)\right) \\ =1-\frac{3}{4}{\left(a^2-1\right)}^2=1-\frac{3}{4}\left(a^4+12a^2\right) \\ =1-\left[\frac{3}{4}{a}^4+\frac{3}{4}-\frac{3}{2}{a}^2\right] \\ =1-\frac{3}{4}-\frac{3}{4}{a}^4+\frac{3}{2}{a}^2 \\ =\frac{1}{4}=\frac{3}{4}{a}^4+\frac{3}{2}{a}^2 \end{gathered}$$
Hence, sin⁶x + cos⁶x