Question

НОМЕР 745 СРОЧНО ДАМ 150 балов

Answer 1

Объяснение:

а) sin²(5π/4-2α)-sin²(5π/4+2α)=

=(sin(5π/4-2α)-sin(5π/4+2α))(sin(5π/4-2α)+sin(5π/4+2α)); подставляем на суммы тригонометрических функций;

$2sin(\frac{\frac{5\pi}{4}-2\alpha-\frac{5\pi}{4}-2\alpha}{2})cos(\frac{\frac{5\pi}{4}-2\alpha+\frac{5\pi}{4}+2\alpha}{2})*2sin(\frac{\frac{5\pi}{4}-2\alpha+\frac{5\pi}{4}+2\alpha}{2})cos(\frac{\frac{5\pi}{4}-2\alpha-\frac{5\pi}{4}-2\alpha}{2})$=

=$4sin(-2\alpha )cos(\frac{5\pi}{4})sin(\frac{5\pi}{4})cos(-2\alpha)=-2sin2\alpha cos2\alpha=-sin4\alpha$

Ответ: -sin4α

б) cos²(3π/8-α/2)-cos²(11π/8+α/2)=(cos(3π/8-α/2)-cos(11π/8+α/2))(cos(3π/8-α/2)+cos(11π/8+α/2))=

=$-2sin(\frac{\frac{3\pi}{8}-\frac{\alpha }{2}+\frac{11\pi}{8}+\frac{\alpha }{2}}{2})sin(\frac{\frac{3\pi}{8}-\frac{\alpha }{2}-\frac{11\pi}{8}-\frac{\alpha }{2}}{2})*2cos(\frac{\frac{3\pi}{8}-\frac{\alpha }{2}+\frac{11\pi}{8}+\frac{\alpha }{2}}{2})cos(\frac{\frac{3\pi}{8}-\frac{\alpha }{2}-\frac{11\pi}{8}-\frac{\alpha }{2}}{2})$=

=-4sin(-7π/8)sin(-π/2-α/2)cos(-7π/8)cos(-π/2-α/2)=4sin(7π/8)cos(α/2)cos(7π/8)sin(α/2)=sin(7π/4)sinα=-√2/2sinα

Ответ: -√2/2sinα