Монгол Бодлогын Сан

Заавар: $\sin^2+\cos^2\alpha=1$ ба $\sin^2\alpha\cos^2\alpha$-аар илэрхийл.

Бодолт: $$\sin^4x+\cos^4x-1=(\sin^2+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha-1=$$ $$=-2\sin^2\alpha\cos^2\alpha$$ ба $$\sin^6x+\cos^6x-1=(\sin^2+\cos^2\alpha)^3-3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)-1=$$ $$=-3\sin^2\alpha\cos^2\alpha$$ тул $\dfrac{\sin^4\alpha+\cos^4\alpha-1}{\sin^6\alpha+\cos^6\alpha-1}=\dfrac{2}{3}$ байна.