$$ \begin{matrix} \sin^2 \alpha + \cos^2 \alpha = 1 & \quad & \mathrm{tg} \, \alpha \cdot \mathrm{ctg} \, \alpha = 1 \\[1em] \mathrm{tg} \, \alpha = \cfrac{\sin \alpha}{\cos \alpha} & \quad & \mathrm{ctg} \, \alpha = \cfrac{\cos \alpha}{\sin \alpha} \\[1em] 1 + \mathrm{tg}^2 \, \alpha = \cfrac{1}{\cos^2 \alpha} & \quad & 1 + \mathrm{ctg}^2 \, \alpha = \cfrac{1}{\sin^2 \alpha} \end{matrix} $$
$$ \sin ( \alpha \pm \beta) = \sin \alpha \cdot \cos \beta \pm \cos \alpha \cdot \sin \beta $$ $$ \cos ( \alpha \pm \beta) = \cos \alpha \cdot \cos \beta \mp \sin \alpha \cdot \sin \beta $$ $$ \mathrm{tg} \, (\alpha \pm \beta ) = \frac{\mathrm{tg} \, \alpha \pm \mathrm{tg} \, \beta}{1 \mp \mathrm{tg} \, \alpha \cdot \mathrm{tg} \, \beta} $$ $$ \mathrm{ctg} \, (\alpha \pm \beta ) = \frac{\mathrm{ctg} \, \alpha \cdot \mathrm{ctg} \, \beta \mp 1}{\mathrm{ctg} \, \beta \pm \mathrm{ctg} \, \alpha} $$
$$ \begin{matrix} \sin 2 \alpha = 2 \sin \alpha \cdot \cos \alpha & \quad & \sin \alpha \cdot \cos \alpha = \cfrac{1}{2} \sin 2 \alpha \\[1em] \cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha & \quad & \\[1em] \cos 2 \alpha = 1 - 2 \sin^2 \alpha & \quad & \cos 2 \alpha = 2 \cos^2 \alpha - 1 \\[1em] \mathrm{tg} \, 2 \alpha = \cfrac{2 \, \mathrm{tg} \, \alpha}{1 - \mathrm{tg}^2 \, \alpha} & \quad & \mathrm{ctg} \, 2 \alpha = \cfrac{\mathrm{ctg}^2 \, \alpha - 1}{2 \, \mathrm{ctg} \, \alpha} \\[1em] \sin 3 \alpha = 3 \sin \alpha - 4 \sin^3 \alpha & \quad & \cos 3 \alpha = 4 \cos^3 \alpha - 3 \cos \alpha \\[1em] \mathrm{tg} \, 3 \alpha = \cfrac{3 \, \mathrm{tg} \, \alpha - \mathrm{tg}^3 \, \alpha}{1 - 3 \,\mathrm{tg}^2 \, \alpha} & \quad & \mathrm{ctg} \, 3 \alpha = \cfrac{\mathrm{ctg}^3 \, \alpha - 3 \, \mathrm{ctg} \, \alpha}{3 \, \mathrm{ctg}^2 \, \alpha - 1} \end{matrix} $$
\begin{matrix} 2 \sin^2 \cfrac{\alpha}{2} = 1 - \cos \alpha & \quad & 2 \cos^2 \cfrac{\alpha}{2} = 1 + \cos \alpha \\[1em] 1 \pm \sin \alpha = \left( \sin \cfrac{\alpha}{2} \pm \cos \cfrac{\alpha}{2} \right)^2 & \quad & \mathrm{tg}^2 \, \cfrac{\alpha}{2} = \cfrac{1 - \cos \alpha}{1 + \cos \alpha} \\[1em] \mathrm{tg} \, \cfrac{\alpha}{2} = \cfrac{\sin \alpha}{1 + \cos \alpha} = \cfrac{1 - \cos \alpha}{\sin \alpha} & & \end{matrix}
$$ \sin \alpha + \sin \beta = 2 \sin \cfrac{\alpha + \beta}{2} \cos \cfrac{\alpha - \beta}{2} \\[1em] \sin \alpha - \sin \beta = 2 \cos \cfrac{\alpha + \beta}{2} \sin \cfrac{\alpha - \beta}{2} \\[1em] \cos \alpha + \cos \beta = 2 \cos \cfrac{\alpha + \beta}{2} \cos \cfrac{\alpha - \beta}{2} \\[1em] \cos \alpha - \cos \beta = -2 \sin \cfrac{\alpha + \beta}{2} \sin \cfrac{\alpha - \beta}{2} \\[1em] \mathrm{tg} \, \alpha \pm \mathrm{tg} \, \beta = \frac{\sin (\alpha \pm \beta)}{\cos \alpha \cos \beta} \\[1em] \mathrm{ctg} \, \alpha \pm \mathrm{ctg} \, \beta = \frac{\sin (\beta \pm \alpha)}{\sin \alpha \sin \beta} \\[1em] \mathrm{tg} \, \alpha \pm \mathrm{ctg} \, \beta = \frac{\pm \cos (\alpha \mp \beta)}{\cos \alpha \sin \beta} $$
$$ \sin \alpha \sin \beta = \frac{1}{2} \left( \cos (\alpha - \beta) - \cos (\alpha + \beta) \right) \\[1em] \cos \alpha \cos \beta = \frac{1}{2} \left( \cos (\alpha - \beta) + \cos (\alpha + \beta) \right) \\[1em] \sin \alpha \cos \beta = \frac{1}{2} \left( \sin (\alpha - \beta) + \sin (\alpha + \beta) \right) $$
Эффективно готовьтесь к ЕГЭ по информатике с новым тренажёром, эмулирующем работу станции КЕГЭ, которая используется на реальном экзамене