668. Упростить: 1) [sin (3π/2 + α) / ctg (2π − α)] · [tg (π/2 + α) / sin (π + α)]; 2) [sin² (π + α) + sin² (π/2 + α)] / cos (3π/2 + α) · ctg (3π/2 − α).

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Accepted Answer

\(\sin(\frac{3\pi}2+\alpha)=-\cos\alpha\), \(\operatorname{ctg}(2\pi-\alpha)=-\operatorname{ctg}\alpha\), \(\operatorname{tg}(\frac\pi2+\alpha)=-\operatorname{ctg}\alpha\), \(\sin(\pi+\alpha)=-\sin\alpha\). Подставляем и сокращаем: \[\frac{-\cos\alpha}{-\operatorname{ctg}\alpha}\cdot\frac{-\operatorname{ctg}\alpha}{-\sin\alpha}=\operatorname{ctg}\alpha.\]
В числителе \(\sin^2(\pi+\alpha)+\sin^2(\frac\pi2+\alpha)=\sin^2\alpha+\cos^2\alpha=1\). Кроме того, \(\cos(\frac{3\pi}2+\alpha)=\sin\alpha\) и \(\operatorname{ctg}(\frac{3\pi}2-\alpha)=\operatorname{tg}\alpha\). Следовательно, \[\frac{1}{\sin\alpha}\cdot\operatorname{tg}\alpha=\frac{1}{\cos\alpha}.\]

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