Упр.679 ГДЗ Никольский Потапов 9 класс (Алгебра)
a) tgа/2=(1-cosа)/sinа
6) tgа/2=sinа/(1+cosа )
в) sin 2a (sin 2a + sin 2b) + cos 2a (cos 2a + cos 2b) = 2 cos^2 (a — b);
г) sin 2a (sin 2a — sin 2b) + cos 2a (cos 2a — cos 2b) = 2 sin^2 (a — b);
д) cos^3 a sin a — sin^3 a cos a = — sin 4a;
е) 2 sin 2a sin a + cos 3a = cosa;
ж) 1 + 2 cos 2a + cos 4a = 4 cos^2 a cos 2a;
з) 1 + 2 cos 3a + cos 6a = 4 cos^2 (3a/2)cos 3a;
и) sin 3a = 3 sin a — 4 sin^3 a;
к) cos 3a = 4 cos^3 a — 3 cos a.
$$\tg \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}$$
$$\frac{1-\cos \alpha}{\sin \alpha} =\frac{2\sin^2 \frac{\alpha}{2}}{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}} =\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} =\tg \frac{\alpha}{2}$$
$$\tg \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}$$
$$\frac{\sin \alpha}{1+\cos \alpha} =\frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}} =\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} =\tg \frac{\alpha}{2}$$
$$\sin 2\alpha(\sin 2\alpha+\sin 2\beta)+\cos 2\alpha(\cos 2\alpha+\cos 2\beta)=2\cos^2(\alpha-\beta)$$
$$\sin^2 2\alpha+\sin 2\alpha\sin 2\beta+\cos^2 2\alpha+\cos 2\alpha\cos 2\beta$$
$$=(\sin^2 2\alpha+\cos^2 2\alpha)+(\sin 2\alpha\sin 2\beta+\cos 2\alpha\cos 2\beta)$$
$$=1+\cos(2\alpha-2\beta)=1+\cos 2(\alpha-\beta)$$
$$=1+\bigl(2\cos^2(\alpha-\beta)-1\bigr)=2\cos^2(\alpha-\beta)$$$$\sin 2\alpha(\sin 2\alpha-\sin 2\beta)+\cos 2\alpha(\cos 2\alpha-\cos 2\beta)=2\sin^2(\alpha-\beta)$$
$$\sin^2 2\alpha-\sin 2\alpha\sin 2\beta+\cos^2 2\alpha-\cos 2\alpha\cos 2\beta$$
$$=(\sin^2 2\alpha+\cos^2 2\alpha)-(\sin 2\alpha\sin 2\beta+\cos 2\alpha\cos 2\beta)$$
$$=1-\cos(2\alpha-2\beta)=1-\cos 2(\alpha-\beta)$$
$$=1-\bigl(1-2\sin^2(\alpha-\beta)\bigr)=2\sin^2(\alpha-\beta)$$$$\cos^3\alpha\sin\alpha-\sin^3\alpha\cos\alpha=\frac14\sin 4\alpha$$
$$\cos^3\alpha\sin\alpha-\sin^3\alpha\cos\alpha =\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$=\frac12\sin 2\alpha\cdot \cos 2\alpha =\frac14\sin 4\alpha$$$$2\sin 2\alpha\sin\alpha+\cos 3\alpha=\cos\alpha$$
$$2\sin 2\alpha\sin\alpha+\cos 3\alpha =2\cdot(2\sin\alpha\cos\alpha)\sin\alpha+\bigl(4\cos^3\alpha-3\cos\alpha\bigr)$$
$$=4\sin^2\alpha\cos\alpha+4\cos^3\alpha-3\cos\alpha$$
$$=\cos\alpha\bigl(4\sin^2\alpha+4\cos^2\alpha-3\bigr)=\cos\alpha$$$$1+2\cos 2\alpha+\cos 4\alpha=4\cos^2\alpha\cos 2\alpha$$
$$1+2\cos 2\alpha+\cos 4\alpha =1+2\cos 2\alpha+\bigl(2\cos^2 2\alpha-1\bigr)$$
$$=2\cos 2\alpha(1+\cos 2\alpha) =2\cos 2\alpha\cdot 2\cos^2\alpha$$
$$=4\cos^2\alpha\cos 2\alpha$$$$1+2\cos 3\alpha+\cos 6\alpha=4\cos^2\frac{3\alpha}{2}\cos 3\alpha$$
$$1+2\cos 3\alpha+\cos 6\alpha =1+2\cos 3\alpha+\bigl(2\cos^2 3\alpha-1\bigr)$$
$$=2\cos 3\alpha(1+\cos 3\alpha) =2\cos 3\alpha\cdot 2\cos^2\frac{3\alpha}{2}$$
$$=4\cos^2\frac{3\alpha}{2}\cos 3\alpha$$$$\sin 3\alpha=3\sin\alpha-4\sin^3\alpha$$
$$\sin 3\alpha=\sin(2\alpha+\alpha)=\sin 2\alpha\cos\alpha+\cos 2\alpha\sin\alpha$$
$$=2\sin\alpha\cos^2\alpha+(1-2\sin^2\alpha)\sin\alpha$$
$$=2\sin\alpha(1-\sin^2\alpha)+\sin\alpha-2\sin^3\alpha$$
$$=3\sin\alpha-4\sin^3\alpha$$$$\cos 3\alpha=4\cos^3\alpha-3\cos\alpha$$
$$\cos 3\alpha=\cos(2\alpha+\alpha)=\cos 2\alpha\cos\alpha-\sin 2\alpha\sin\alpha$$
$$=(\cos^2\alpha-\sin^2\alpha)\cos\alpha-2\sin^2\alpha\cos\alpha$$
$$=\cos^3\alpha-3\sin^2\alpha\cos\alpha$$
$$=\cos^3\alpha-3(1-\cos^2\alpha)\cos\alpha$$
$$=4\cos^3\alpha-3\cos\alpha$$
Ответ
Все равенства доказаны.
