Question

!!!СРОЧНО!!!
ПОМОГИТЕ ПОЖАЛУЙСТА
ДАМ 50 БАЛЛОВ

Answer 1

Ответ:

1.

2

$150^{\circ} = 90^{\circ} + \alpha \\ \alpha = 150^{\circ} - 90^{\circ} = 60^{\circ}$

4

$310^{\circ} = 270^{\circ} + \alpha \\ \alpha = 310^{\circ} - 270^{\circ} = 40^{\circ}$

6

$\frac{\pi}{5} = \frac{\pi}{2} - \alpha \\ \alpha = \frac{\pi}{2} - \frac{\pi}{5} = \frac{5 - 2}{10} \pi= \frac{3\pi}{10}$

8

$\frac{11\pi}{6} = 2\pi - \alpha \\ \alpha = 2\pi - \frac{11\pi}{6} = \frac{\pi}{6}$

2.

5

$\cos(225^{\circ}) = \cos(180^{\circ} + 45^{\circ}) = - \cos(45^{\circ}) = - \frac{ \sqrt{2} }{2}$

6

$\sin(210^{\circ}) = \sin(180^{\circ} + 30^{\circ}) = - \sin(30^{\circ}) = - \frac{1}{2}$

7

$ctg(240^{\circ}) = ctg(270^{\circ} - 30^{\circ}) = tg(30^{\circ}) = \frac{ \sqrt{3} }{3}$

8

$\sin(315^{\circ}) = \sin(360^{\circ} - 45^{\circ}) = - \sin(45^{\circ}) = - \frac{ \sqrt{2} }{2}$

1

$tg \frac{5\pi}{4} = tg(\pi + \frac{\pi}{4} ) = tg \frac{\pi}{4} = 1$

2

$\sin( \frac{7\pi}{6} ) = \sin( \pi+ \frac{\pi}{6} ) = - \sin( \frac{\pi}{6} ) = - \frac{1}{2}$

3

$\cos( \frac{5\pi}{3} ) = \cos( 2\pi - \frac{\pi}{3} ) = \cos( \frac{\pi}{3} ) = \frac{1}{2}$

4

$ctg \frac{5\pi}{3} = ctg(2\pi - \frac{\pi}{3} ) - ctg \frac{\pi}{3} = - \frac{ \sqrt{3} }{3}$

3.

2

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

4

$\cos {}^{2} ( \alpha - \frac{\pi}{4} ) - \cos {}^{2} ( \alpha + \frac{\pi}{4} ) = \\ = {( \cos( \alpha ) \cos( \frac{\pi}{4} ) + \sin( \alpha ) \sin( \frac{\pi}{4} ) )}^{2} - {( \cos( \alpha ) \cos( \frac{\pi}{4} ) - \sin( \alpha ) \sin( \frac{\pi}{4} ) )}^{2} = \\ = {( \frac{ \sqrt{2} }{2} \cos( \alpha ) + \frac{ \sqrt{2} }{2} \sin( \alpha ) )}^{2} - {( \frac{ \sqrt{2} }{2} \cos( \alpha ) - \frac{ \sqrt{2} }{2} \sin( \alpha ) )}^{2} = \\ = \frac{1}{2} {( \cos( \alpha ) + \sin( \alpha ) )}^{2} - \frac{1}{2} {( \cos( \alpha ) - \sin( \alpha ) )}^{2} = \\ = \frac{1}{2} ( \cos( \alpha ) + \sin( \alpha ) - \cos( \alpha ) + \sin( \alpha ) )( \cos( \alpha ) + \sin( \alpha ) + \cos( \alpha ) - \sin( \alpha )) = \\ = \frac{1}{2} \times 2 \sin( \alpha ) \times 2 \cos( \alpha ) = \sin(2 \alpha )$

4.

2

$\sin(105^{\circ}) - \sin(75^{\circ}) = \\ = 2 \sin( \frac{105^{\circ} - 75^{\circ}}{2} ) \cos( \frac{105^{\circ} + 75^{\circ}}{2} ) = \\ = 2 \sin(15^{\circ}) \cos(90^{\circ}) = 0$

4

$\cos( \frac{11\pi}{12} ) - \cos( \frac{5\pi}{12} ) = \\ = - 2 \sin( \frac{1}{2}( \frac{11\pi}{12} - \frac{5\pi}{12} ) ) \sin( \frac{1}{2} ( \frac{11\pi}{12} + \frac{5\pi}{12} ) ) = \\ = - 2 \sin( \frac{\pi}{4} ) \times \sin( \frac{2\pi}{3} ) = - 2 \times \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} = - \frac{ \sqrt{6} }{2}$

6

$\sin(105^{\circ}) + \sin(165^{\circ}) = \\ = 2 \sin( \frac{105^{\circ} + 165^{\circ}}{2} ) \cos( \frac{105^{\circ} - 165^{\circ}}{2} ) = \\ = 2 \sin(135^{\circ}) \cos(30^{\circ}) = 2 \sin(180^{\circ} - 45^{\circ}) \cos(30^{\circ}) = \\ = 2 \sin(45^{\circ}) \times \frac{ \sqrt{3} }{2} = 2 \times \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} = \frac{ \sqrt{6} }{2}$