Упр.4.8 ГДЗ Мордкович Семенов 10 класс (Алгебра)
а) $$\cos\left(-\frac{3\pi}{4}\right)+\cos\left(\frac{\pi}{4}\right)+\cos\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}\right)+\cos(0)\sin\left(\frac{\pi}{2}\right)$$
б) $$\sin\left(-\frac{2\pi}{3}\right)-\cos\left(\frac{3\pi}{4}\right)+\cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{6}\right)+\cos\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)$$
в) $$\cos\left(-\frac{\pi}{4}\right)+\sin^2\left(\frac{\pi}{3}\right)+\sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{2}\right)-\cos\left(\frac{5\pi}{4}\right)$$
г) $$\sin\left(\frac{5\pi}{6}\right)+\cos\left(\frac{3\pi}{4}\right)+\cos\left(\frac{\pi}{4}\right)\sin\left(\frac{3\pi}{4}\right)+\cos\left(\frac{\pi}{2}\right)\sin(\pi)$$
а) $$\cos\left(-\frac{3\pi}{4}\right)+\cos\frac{\pi}{4}+\cos\frac{\pi}{6}\cdot\sin\frac{\pi}{6}+\cos 0\cdot\sin\frac{\pi}{2}$$
$$=-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}+\frac{\sqrt3}{2}\cdot\frac12+1\cdot1$$
$$=\frac{\sqrt3}{4}+1$$
б) $$\sin\left(-\frac{2\pi}{3}\right)-\cos\frac{3\pi}{4}+\cos\frac{\pi}{3}\cdot\sin\frac{\pi}{6}+\cos\frac{\pi}{2}\cdot\sin\frac{\pi}{2}$$
$$=-\frac{\sqrt3}{2}-\left(-\frac{\sqrt2}{2}\right)+\frac12\cdot\frac12+0\cdot1$$
$$=\frac{-2\sqrt3+2\sqrt2+1}{4}$$
в) $$\cos\left(-\frac{\pi}{4}\right)+\sin^2\frac{\pi}{3}+\sin\frac{\pi}{6}\cdot\sin\frac{\pi}{2}-\cos\frac{5\pi}{4}$$
$$=\frac{\sqrt2}{2}+\left(\frac{\sqrt3}{2}\right)^2+\frac12\cdot1-\left(-\frac{\sqrt2}{2}\right)$$
$$=\frac{\sqrt2}{2}+\frac34+\frac12+\frac{\sqrt2}{2}=\sqrt2+\frac54=\frac{4\sqrt2+5}{4}$$
г) $$\sin\frac{5\pi}{6}+\cos\frac{3\pi}{4}+\cos\frac{\pi}{4}\cdot\sin\frac{3\pi}{4}+\cos\frac{\pi}{2}\cdot\sin\pi$$
$$=\frac12-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}+0\cdot0$$
$$=\frac12-\frac{\sqrt2}{2}+\frac12=1-\frac{\sqrt2}{2}=\frac{2-\sqrt2}{2}$$
Ответ
а) $$1+\frac{\sqrt3}{4}$$; б) $$\frac{-2\sqrt3+2\sqrt2+1}{4}$$; в) $$\frac{4\sqrt2+5}{4}$$; г) $$\frac{2-\sqrt2}{2}$$.
