Монгол Бодлогын Сан

Заавар:

Бодолт: $$\begin{align*} \sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^22\alpha&=\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}(2\sin\alpha\cos\alpha)^2\\ &=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cos^2\alpha\\ &=(\sin^2\alpha+\cos^2\alpha)^2=1^2=1\\ \end{align*}$$