Монгол Бодлогын Сан

Заавар: Эмхэтгэлийн томьёо ашиглан $[0^\circ;90^\circ]$ өнцгөөр илэрхийл.

Бодолт: $$\begin{align*} \dfrac{2\sin^270^{\circ}-1}{2\ctg115^{\circ}\cdot\cos^2155^{\circ}} & =\dfrac{–\cos 140^\circ}{2\tg 25^\circ\cdot(-\sin 65^\circ)^2}\\ &=\dfrac{\sin 50^\circ}{2\cdot\frac{\sin 25^\circ}{\cos 25^\circ}\cdot\sin^265^\circ} & &\color{red}{\leftarrow \cos(90^\circ+\alpha)=-\sin\alpha}\\ &=-\dfrac{2\sin 25^\circ\cos^2 25^\circ}{2\sin 25^\circ\sin^265^\circ} & &\color{red}{\leftarrow \sin2\alpha=2\sin\alpha\cos\alpha}\\ &=-\dfrac{\cos^225^\circ}{\sin^265^\circ}=-1 & & \color{red}{\leftarrow \sin65^\circ=\sin(90^\circ-25^\circ)=\cos25^\circ} \end{align*}$$