Question

помогите кто может. ​

Answer 1

Ответ:

1.

формула косинуса двойного угла:

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

$2 \sin {}^{2} ( \frac{\pi}{4} + \frac{3x}{2} ) - 1 = \\ = - (1 - 2 \sin {}^{2} ( \frac{\pi}{4} + \frac{3x}{2} ) ) = \\ = - \cos(2( \frac{\pi}{4} + \frac{3x}{2} )) = \\ = - \cos( \frac{\pi}{2} + 3x ) = - ( - \sin(3x) ) = \\ = \sin(3x)$

2.

$\cos( \alpha ) = \frac{7}{25}$

угол принадлежит 4 четверти, синус отрицательный.

$\sin( \alpha ) = \sqrt{1 - \cos {}^{2} ( \alpha ) } \\ \sin( \alpha ) = - \sqrt{1 - \frac{49}{625} } = - \sqrt{ \frac{576}{625} } = - \frac{24}{25}$

$\sin( 2\alpha ) = 2 \sin( \alpha ) \cos( \alpha ) = \\ = 2 \times ( - \frac{24}{25} ) \times \frac{7}{25} = - \frac{336}{625}$

$\cos( 2\alpha ) = 2 \cos {}^{2} ( \alpha ) - 1 = \\ = 2 \times {( \frac{7}{25}) }^{2} - 1 = 2 \times \frac{49}{625} - 1 = \\ = \frac{98 - 625}{625} = - \frac{527}{625}$

$tg 2\alpha = \frac{ \sin( 2\alpha ) }{ \cos( 2\alpha ) } = \\ = - \frac{336}{625} \times ( - \frac{625}{527} ) = \frac{336}{527}$

3.

$\cos(15°) = \cos(45° - 30°) = \\ = \cos(45°) \cos(30°) + \sin(45°) \sin(30°) = \\ = \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} + \frac{ \sqrt{2} }{2} \times \frac{1}{2} = \\ = \frac{ \sqrt{2} }{2} ( \frac{ \sqrt{3} }{2} + \frac{1}{2} ) = \frac{ \sqrt{2} }{2} \times \frac{1 + \sqrt{3} }{2} = \\ = \frac{ \sqrt{2}(1 + \sqrt{3} )}{4}$