Question

Тригонометрические выражения

Answer 1

Ответ:

1.

a

$\sin(2\pi) - \cos( \frac{3\pi}{2} ) = 0 - 0 = 0$

б

$tg(2\pi) - \cos(2\pi) = 0 - 1 = - 1$

в

$\cos(\pi) + \sin( \frac{3\pi}{2} ) = - 1 - 1 = - 2$

2.

а

$\sin( \frac{\pi}{4} ) + 3 \cos( \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2} + 3 \times \frac{ \sqrt{2} }{2} = \\ = 4 \times \frac{ \sqrt{2} }{2} = 2 \sqrt{2}$

б

$tg \frac{\pi}{6} + \frac{1}{3} \sin( \frac{\pi}{3} ) - \frac{2}{3} \cos( \frac{\pi}{6} ) = \\ = \frac{ \sqrt{3} }{3} + \frac{1}{3} \times \frac{ \sqrt{3} }{2} - \frac{2}{3} \times \frac{ \sqrt{3} }{2} = \\ = \frac{ \sqrt{3} }{3} + \frac{ \sqrt{3} }{6} - \frac{ \sqrt{3} }{3} = \frac{ \sqrt{3} }{6}$

3.

$\cos( \alpha ) = - \frac{12}{13}$

угол принадлежит 3 четверти, синус < 0

$\sin( \alpha ) = \sqrt{1 - \cos {}^{2} ( \alpha ) } \\ \sin( \alpha ) = - \sqrt{1 - \frac{144}{169} } = - \sqrt{ \frac{25}{169} } = - \frac{5}{13}$

$tg \alpha = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = - \frac{5}{13} \times ( - \frac{13}{12} ) = \frac{5}{12}$

$ctg \alpha = \frac{1}{tg \alpha } = \frac{12}{5}$

4.

$6 \cos {}^{2} ( \alpha ) + 6 \sin {}^{2} ( \alpha ) + 5 = \\ = 6( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha )) + 5 = 6 + 5 = 11$

5.

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