Question

Срочнооо!!!! Прошу помогите
Вариант В1

Answer 1

1.

а)

$\frac{sin \frac{3\pi}{4}cos \frac{\pi}{12} + cos\frac{3\pi}{4} sin \frac{\pi}{12} }{ \sin15°} = \\ = \frac{ \sin( \frac{3\pi}{4} + \frac{\pi}{12} ) }{sin45° - 30°} = \\ = \frac{sin \frac{5\pi}{6} }{sin45°cos30° - cos45°sin30°} = \\ = \frac{ \sin(\pi - \frac{\pi}{6} ) }{ \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{2} }{2} \times \frac{1}{2} } = \\ = \frac{ \sin\frac{\pi}{6} }{ \frac{1}{2} ( \sqrt{6} - \sqrt{2} )} = \frac{ \frac{1}{2} }{ \frac{1}{2} ( \sqrt{6} - \sqrt{2}) } = \\ = \frac{1}{ \sqrt{6} - \sqrt{2} } = \frac{ \sqrt{6} + \sqrt{2} }{4}$

б)

$cos \alpha = - \sqrt{ \frac{1}{1 + {tg}^{2} \alpha } } = \\ = - \sqrt{ \frac{1}{1 + \frac{1}{3} } } = - \sqrt{ \frac{3}{4} } = - \frac{ \sqrt{3} }{2} \\ \alpha = arccos( - \frac{ \sqrt{3} }{2}) = \\ = \pi - arccos \frac{ \sqrt{3} }{2} = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

$cos \alpha = - \sqrt{ \frac{1}{1 + {tg}^{2} \alpha } } = \\ = - \sqrt{ \frac{1}{1 + \frac{1}{3} } } = - \sqrt{ \frac{3}{4} } = - \frac{ \sqrt{3} }{2} \\ \alpha = arccos( - \frac{ \sqrt{3} }{2}) = \\ = \pi - arccos \frac{ \sqrt{3} }{2} = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$\cos( \frac{\pi}{6} + \alpha ) = \cos( \frac{\pi}{6} + \frac{5\pi}{6} ) = cos\pi = - 1$

2.

а).${ \cos( \frac{\pi}{12 } + \alpha ) \cos( \frac{ 5\pi}{12 } - \alpha )- \sin( \frac{\pi}{12 } + \alpha ) \sin( \frac{5\pi}{12 } - \alpha )} =0 \\ \cos( \frac{\pi}{12} + \alpha + \frac{5\pi}{12} - \alpha ) = 0 \\ \cos \frac{\pi}{2} = 0 \\ 0 = 0$

б)

$\sin( \alpha - \beta ) \sin( \alpha + \beta ) = { \sin }^{2} \alpha - {sin}^{2} \beta \\( sin \alpha \: cos \beta - cos \alpha \: sin \beta )(sin \alpha \: cos \beta + cos \alpha \: sin \beta) = { \sin }^{2} \alpha - {sin}^{2} \beta \\{ \sin }^{2} \alpha \: {cos}^{2} \beta - {cos}^{2} \alpha \: {sin}^{2} \beta = { \sin }^{2} \alpha - {sin}^{2} \beta \\ {\sin }^{2} \alpha(1 - {\sin }^{2} \beta ) - {\sin }^{2} \beta ( 1 - {\sin }^{2} \alpha) = {\sin }^{2} \alpha - {\sin }^{2} \beta \\ {\sin }^{2} \alpha - {\sin }^{2} \alpha{\sin }^{2} \beta - {\sin }^{2} \beta + {\sin }^{2} \alpha{\sin }^{2} \beta = {\sin }^{2} \alpha - {\sin }^{2} \beta \\ {\sin }^{2} \alpha - {\sin }^{2} \beta= {\sin }^{2} \alpha - {\sin }^{2} \beta$

3.

$ctg \alpha = \frac{1}{3} \\tg( \alpha + \beta ) = 1 \\ tg \alpha = \frac{1}{ctg \alpha } = \frac{1}{ \frac{1}{3} } = 3 \\ tg( \alpha + \beta ) = \frac{tg \alpha + tg \beta }{1 - tg \alpha \: tg \beta } \\ \frac{3 + tg \beta }{1 - 3tg \beta } = 1 \\ 3 + tg \beta = 1 - 3tg \beta \\ 4tg \beta = - 2 \\ tg \beta = - \frac{2}{4} = - \frac{1}{2} \\ ctg \beta = \frac{1}{tg \beta } = \frac{1}{ - \frac{1}{2} } = - 2$