Question

Даю 70 баллов, пожалуйста!!!​

Answer 1

Ответ:

1.

1)

$\cos(42^{\circ}) \cos(18^{\circ}) - \sin(42^{\circ}) \sin(18^{\circ}) = \\ = \cos(42^{\circ} + 18^{\circ}) = \cos(60^{\circ}) = \frac{1}{2}$

2)

$\cos( \frac{8\pi}{7} ) \cos( \frac{\pi}{7} ) + \sin( \frac{8\pi}{7} ) \sin( \frac{\pi}{7} ) = \\ = \cos( \frac{8\pi}{7} - \frac{\pi}{7} ) = \cos(\pi) = - 1$

3)

$\sin(32^{\circ}) \cos(62^{\circ}) - \sin(62) \cos(32^{\circ}) = \\ = \sin(32^{\circ} - 62^{\circ}) = \sin( - 30^{\circ}) = - \frac{1}{2}$

4)

$\frac{tg \frac{\pi}{9} + tg \frac{2\pi}{9} }{1 - tg \frac{\pi}{9} tg \frac{2\pi}{9} } = tg( \frac{\pi}{9} + \frac{2\pi}{9} ) = tg( \frac{\pi}{3} ) = \sqrt{3}$

5)

$2 \sin(15^{\circ}) \cos(15^{\circ}) = \sin(2 \times 15^{\circ}) = \sin(30^{\circ}) = \frac{1}{2}$

6)

$\cos {}^{2} ( \frac{\pi}{12} ) - \sin {}^{2} ( \frac{\pi}{12} ) = \cos(2 \times \frac{\pi}{12} ) \cos( \frac{\pi}{6} ) = \frac{ \sqrt{3} }{2}$

2.

1)

$\cos( \alpha ) = \frac{4}{5}$

угол принадлежит 4 четверти, синус отрицательный

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

$\sin( \alpha + \frac{\pi}{3} ) = \\ = \sin( \alpha ) \cos( \frac{\pi}{3} ) + \cos( \alpha ) \sin( \frac{\pi}{3} ) = \\ = \frac{1}{2} \sin( \alpha ) + \frac{ \sqrt{3} }{2} \cos( \alpha ) = \\ \\ = \frac{1}{2} \times ( - \frac{3}{5} ) + \frac{ \sqrt{3} }{2} \times \frac{4}{5} = \\ = \frac{ - 3 + 4 \sqrt{3} }{10}$

2)

$\sin( \alpha ) = 0.6 \\ \cos( \alpha ) = - \sqrt{1 - {(0.6)}^{2} } = - \sqrt{1 - 0.36} = \\ = - \sqrt{0.64} = - 0.8 \\ \\ \cos( \beta ) = - 0.8 \\ \sin( \beta ) = \sqrt{1 - 0.64} = \sqrt{036} = 0.6$

$\cos( \alpha + \beta ) = \\ = \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) = \\ = - 0.8 \times ( - 0.8) - 0.6 \times0.6 = \\ = 0.64 - 0.36 = 0.28$

3.

1)

$\frac{ \cos(3x) \cos(x) + \sin(3x) \sin(x) }{ 2\sin(x) \cos(x) } = \frac{ \cos(3x - x) }{ \sin(2x) } = \\ = \frac{ \cos(2x) }{ \sin(2x) } = ctg(2x)$

2)

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

4.

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

5.

1)

$\sin( \frac{5\pi}{6} + \alpha ) \cos( \frac{\pi}{3} + \alpha ) - \cos( \frac{5\pi}{6} + \alpha ) \sin( \frac{\pi}{3} + \alpha ) = \\ = \sin( \frac{5\pi}{6} + \alpha - \frac{\pi}{3} - \alpha ) = \sin( \frac{5\pi - 2\pi}{ 3} ) = \sin(\pi) = 0$

2)

$( \frac{ \cos( \alpha ) }{ \sin(3 \alpha ) } + \frac{ \sin( \alpha ) }{ \cos(3 \alpha ) } ) \sin( 6\alpha ) - \cos( 2\alpha ) = \\ = \frac{ \cos( \alpha ) \cos(3 \alpha ) + \sin( \alpha ) \sin(3 \alpha ) }{ \sin( 3\alpha ) \cos( 3\alpha ) } \times 2 \sin( 3\alpha ) \cos(3 \alpha ) - \cos( 2\alpha ) = \\ = \cos( \alpha - 3 \alpha ) \times 2 - \cos(2 \alpha ) = 2 \cos( 2\alpha ) - \cos( 2\alpha ) = \cos(2 \alpha )$

3)

$\frac{ \sin( \alpha - \beta ) - \sin( \beta ) \cos(\pi - \alpha ) }{ \cos( \alpha + \beta ) + \sin(\pi - \beta ) \sin( \alpha ) } = \\ = \frac{ \sin( \alpha ) \cos( \beta ) - \cos( \alpha ) \sin( \beta ) + \sin( \beta ) \cos( \alpha ) }{ \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) + \sin( \alpha ) \sin( \beta ) } = \\ = \frac{ \sin( \alpha ) \cos( \beta ) }{ \cos( \alpha ) \cos( \beta ) } = tg( \alpha )$