Ta có sin^8 x + cos^8 x = a/64 + b/16 cos4x + c/64 cos8x với a,b thuộc Q

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} {\sin }^{8}x+{\cos }^{8}x={\left({\sin }^{4}x+{\cos }^{4}x\right)}^2-2{\sin }^{4}x.{\cos }^{4}x \\ =\left(1-2{\sin }^{2}x.{\cos }^{2}x\right)-\frac{1}{8}{\sin }^{4}2x \\ ={\left(1-\frac{1}{2}{\sin }^{2}2x.\right)}^2-\frac{1}{8}{\sin }^{4}2x=1-{\sin }^{2}2x+\frac{1}{8}{\sin }^{4}2x \\ =1-\frac{1-\cos 4x}{2}+\frac{1}{8}{\left(\frac{1-\cos 4x}{2}\right)}^2 \\ =1-\frac{1-\cos 4x}{2}+\frac{1}{32}{\left(1-2\cos 4x+\frac{1+\cos 8x}{2}\right)}^2 \\ =\frac{35}{64}+\frac{7}{16}\cos 4x+\frac{1}{64}\cos 8x \\ \Rightarrow a=35,b=7,c=1\Rightarrow a-5b+c=1 \end{gathered}$$