Формули подвійного і половинного кута. формули пониження степеня

$\sin 2\alpha = \sin^2\alpha - \cos^2\alpha$

$\sin 2\alpha = 2\sin\alpha\cos\alpha$

$\sin 2\alpha = 2\sin\alpha$

$\sin 2\alpha = \cos\alpha\sin\alpha$

$\cos 2\alpha = \sin^2\alpha - \cos^2\alpha$

$\cos 2\alpha = 1 - \sin^2\alpha$

$\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$

$\cos 2\alpha = 2\cos^2\alpha + 1$

$\text{tg } 2\alpha = \frac{\text{tg }\alpha}{1 - \text{tg }^2\alpha}$

$\text{tg } 2\alpha = \frac{2\text{tg }\alpha}{1 + \text{tg }^2\alpha}$

$\text{tg } 2\alpha = 2\text{tg }\alpha$

$\text{tg } 2\alpha = \frac{2\text{tg }\alpha}{1 - \text{tg }^2\alpha}$

$\sin^2\alpha = \frac{1 - \cos 2\alpha}{2}$

$\sin^2\alpha = \frac{1 + \cos 2\alpha}{2}$

$\sin^2\alpha = 1 - \cos 2\alpha$

$\sin^2\alpha = \frac{1 - \sin 2\alpha}{2}$

$\cos x$

$-\sin x$

$\sin x$

1

$\sin 70^\circ + \cos 70^\circ$

$\cos 70^\circ - \sin 70^\circ$

1

$-(\sin 70^\circ + \cos 70^\circ)$

$\frac{\sqrt{3}}{2}$

$-\frac{1}{2}$

$-\frac{\sqrt{3}}{2}$

$\frac{1}{2}$

$\sin 4\alpha$

$2\sin 8\alpha$

$\cos 8\alpha$

$\sin 8\alpha$

$2 \text{tg } 20^\circ$

$2 \text{tg } 40^\circ$

$4 \text{tg } 40^\circ$

$\text{tg } 40^\circ$

$\text{tg } 4x$

$\text{tg } 2x$

$\text{ctg } 4x$

1

$\text{tg }^2\alpha$

$2\text{tg } 2\alpha$

$\text{tg } 2\alpha$

0