Упр.19.3 ГДЗ Мордкович 10-11 класс (Алгебра)
- а) $$\sin\left(\frac{5\pi}{6}-\alpha\right)-\frac{1}{2}\cos\alpha$$;
б) $$\sqrt{3}\cos\alpha-2\cos\left(\alpha-\frac{\pi}{6}\right)$$;
в) $$\frac{\sqrt{3}}{2}\sin\alpha+\cos\left(\alpha-\frac{5\pi}{3}\right)$$;
г) $$\sqrt{2}\sin\left(\alpha-\frac{\pi}{4}\right)-\sin\alpha$$.
а) $$\sin\left(\frac{5\pi}{6}-a\right)-\frac12\cos a$$
$$\sin\left(\frac{5\pi}{6}-a\right)-\frac12\cos a = \sin\frac{5\pi}{6}\cos a-\cos\frac{5\pi}{6}\sin a-\frac12\cos a$$
$$= \frac12\cos a+\frac{\sqrt3}{2}\sin a-\frac12\cos a = \frac{\sqrt3}{2}\sin a$$
Ответ: $$\frac{\sqrt3}{2}\sin a$$
б) $$\sqrt3\cos a-2\cos\left(a-\frac{\pi}{6}\right)$$
$$\sqrt3\cos a-2\cos\left(a-\frac{\pi}{6}\right) = \sqrt3\cos a-2\left(\cos a\cos\frac{\pi}{6}+\sin a\sin\frac{\pi}{6}\right)$$
$$= \sqrt3\cos a-2\left(\frac{\sqrt3}{2}\cos a+\frac12\sin a\right) = \sqrt3\cos a-\sqrt3\cos a-\sin a = -\sin a$$
Ответ: $$-\sin a$$
в) $$\frac{\sqrt3}{2}\sin a+\cos\left(a-\frac{5\pi}{3}\right)$$
$$\frac{\sqrt3}{2}\sin a+\cos\left(a-\frac{5\pi}{3}\right) = \frac{\sqrt3}{2}\sin a+\cos\left(a+\frac{\pi}{3}\right)$$
$$= \frac{\sqrt3}{2}\sin a+\left(\cos a\cos\frac{\pi}{3}-\sin a\sin\frac{\pi}{3}\right)$$
$$= \frac{\sqrt3}{2}\sin a+\frac12\cos a-\frac{\sqrt3}{2}\sin a = \frac12\cos a$$
Ответ: $$\frac12\cos a$$
г) $$\sqrt2\sin\left(a-\frac{\pi}{4}\right)-\sin a$$
$$\sqrt2\sin\left(a-\frac{\pi}{4}\right)-\sin a = \sqrt2\left(\sin a\cos\frac{\pi}{4}-\cos a\sin\frac{\pi}{4}\right)-\sin a$$
$$= \sqrt2\left(\sin a\cdot\frac{\sqrt2}{2}-\cos a\cdot\frac{\sqrt2}{2}\right)-\sin a = \sin a-\cos a-\sin a = -\cos a$$
Ответ: $$-\cos a$$
