Упр.539 ГДЗ Алимов 10-11 класс (Алгебра)
Преобразовать в произведение:
- $$1+2\sin\alpha$$;
- $$1-2\sin\alpha$$;
- $$1+2\cos\alpha$$;
- $$1+\sin\alpha$$.
$$1+2\sin \alpha=2\left(\frac12+\sin \alpha\right)=2(\sin 30^\circ+\sin \alpha)$$
$$=4\sin\frac{30^\circ+\alpha}{2}\cos\frac{30^\circ-\alpha}{2}=4\sin\left(15^\circ+\frac{\alpha}{2}\right)\cos\left(15^\circ-\frac{\alpha}{2}\right).$$
$$1-2\sin \alpha=2\left(\frac12-\sin \alpha\right)=2(\sin 30^\circ-\sin \alpha)$$
$$=4\sin\frac{30^\circ-\alpha}{2}\cos\frac{30^\circ+\alpha}{2}=4\sin\left(15^\circ-\frac{\alpha}{2}\right)\cos\left(15^\circ+\frac{\alpha}{2}\right).$$
$$1+2\cos \alpha=2\left(\frac12+\cos \alpha\right)=2(\cos 60^\circ+\cos \alpha)$$
$$=4\cos\frac{60^\circ+\alpha}{2}\cos\frac{60^\circ-\alpha}{2}=4\cos\left(30^\circ+\frac{\alpha}{2}\right)\cos\left(30^\circ-\frac{\alpha}{2}\right).$$
$$1+\sin \alpha=\sin 90^\circ+\sin \alpha=2\sin\frac{90^\circ+\alpha}{2}\cos\frac{90^\circ-\alpha}{2}$$
$$=2\sin\left(45^\circ+\frac{\alpha}{2}\right)\cos\left(45^\circ-\frac{\alpha}{2}\right).$$









