Question

Помогите с решением.

Answer 1

1) Так как $\frac{\pi}{2} < \alpha < \pi$, то cos α < 0.

По основному тригонометрическому тождеству:

$sin^2\alpha + cos^2\alpha = 1\\cos^2\alpha = 1 - sin^2\alpha = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}\\cos\alpha = -\frac{3}{5}$

По формуле косинуса суммы:

$cos(60^o + \alpha) = cos60^{o}cos\alpha - sin60^{o}sin\alpha = \frac{1}{2} * (-\frac{3}{5}) - \frac{\sqrt{3}}{2}*\frac{4}{5} = -\frac{3}{10} - \frac{2\sqrt{3}}{5} = -0,3 - 0,4\sqrt{3}.$

Ответ: -0,3 - 0,4√3.

2)

$\frac{sin\frac{\pi}{10}*sin\frac{\pi}{5} + cos\frac{\pi}{10}*cos\frac{\pi}{5}}{sin\frac{\pi}{5}*sin\frac{2\pi}{15} - cos\frac{\pi}{5} * cos\frac{2\pi}{15}} = \frac{cos(\frac{\pi}{10} - \frac{\pi}{5})}{-(cos\frac{\pi}{5}*cos\frac{2\pi}{15} - sin\frac{\pi}{5}*sin\frac{2\pi}{15})} = \frac{cos(-\frac{\pi}{10})}{-cos(\frac{\pi}{5}+\frac{2\pi}{15})} = \frac{cos\frac{\pi}{10}}{-cos\frac{\pi}{3}} =\\\\= -\frac{cos\frac{\pi}{10}}{\frac{1}{2}} = -2cos\frac{\pi}{10}.$

Ответ: $-2cos\frac{\pi}{10}$.

3)

$\frac{sin(\alpha -\beta)}{tg\alpha - tg\beta} = cos\alpha*cos\beta\\\\\frac{sin(\alpha -\beta)}{\frac{sin\alpha}{cos\alpha} - \frac{sin\beta}{cos\beta}} = cos\alpha*cos\beta\\\\\frac{sin(\alpha-\beta)}{\frac{sin\alpha cos\beta - sin\beta cos\alpha}{cos\alpha cos\beta}} = cos\alpha*cos\beta\\\\\frac{sin(\alpha - \beta)cos\alpha cos\beta}{sin(\alpha - \beta)} = cos\alpha * cos\beta\\\\cos\alpha cos\beta = cos\alpha*cos\beta$, ч.т.д.