Монгол Бодлогын Сан

Заавар: $$\cos2x=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1$$

Бодолт: $$\begin{align*} \sin^8x-\cos^8x&=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)(\sin^4x+\cos^4x)\\ &=-\cos2x\cdot 1\cdot\bigg\{\Big(\dfrac{1-\cos2x}{2}\Big)^2+\Big(\dfrac{1+\cos2x}{2}\Big)^2\bigg\}\\ &=\dfrac{3}{5}\cdot\bigg\{\Big(\dfrac{1+\frac35}{2}\Big)^2+\Big(\dfrac{1-\frac35}{2}\Big)^2\bigg\}\\ &=\dfrac{3}{5}\cdot\bigg\{\Big(\dfrac{4}{5}\Big)^2+\Big(\dfrac{1}{5}\Big)^2\bigg\}=\dfrac{51}{125} \end{align*}$$