Question

Помогите, пожалуйста.
1. Найдите значение выражения:
cos240°·cos225°·sin135°+cos120°

2. Упростите
cos(3π/2+α)·tg(π/2-α)-sin(π/2-α)+ctg(3π/2-α)-ctg(π/2-α)

3. Упростите левую часть выражения и покажите на единичной окружности угол x:
cos(π/2-x)+sin(2π+x)=2

4. Докажите тождество:
sin(π-α)ctg(π/2-α)cos(2π-α)/tg(π+α)tg(π/2+α)sin(-α)=sinα

Answer 1

1)
$\displaystyle cos(240^o)*cos(225^o)*sin(135^o)+cos(120^o) = cos(180^o+60^o)*cos(180^o+45^o)*sin(180^o-45^o)+cos(90^o+30^o) = -cos(60^o)*(-cos(45^o))*sin(45^o)-sin(30^o) = -\frac{1}{2} *(-\frac{\sqrt{2} }{2} )*\frac{\sqrt{2} }{2} -\frac{1}{2} =\frac{2}{8}-\frac{1}{2}=\frac{2-4}{8} =-\frac{2}{8}=-\frac{1}{4}=-0,25$

2)
$\displaystyle cos(\frac{3\pi }{2}+\alpha )*tg(\frac{\pi }{2}-\alpha )-sin(\frac{\pi }{2}-\alpha )+ctg(\frac{3\pi }{2}-\alpha )-ctg(\frac{\pi }{2}-\alpha )=sin(\alpha )*ctg(\alpha )-cos(\alpha )+tg(\alpha )=sin(\alpha )*\frac{cos(\alpha )}{sin(\alpha )} -cos(\alpha )+tg(\alpha )=cos(\alpha )-cos(\alpha )+tg(\alpha )=tg(\alpha )$

3) $\displaystyle cos(\frac{\pi }{2}-x)+sin(2\pi +x)=2$
$\displaystyle sin(x)+sin(x)=2$
$\displaystyle 2sin(x)=2|:2$
$\displaystyle sin(x)=1$
(далее см. вложение)

4)
$\displaystyle \frac{sin(\pi -\alpha )*ctg(\frac{\pi }{2}-\alpha )*cos(2\pi -\alpha )}{tg(\pi +\alpha )*tg(\frac{\pi }{2} +\alpha )*sin(-\alpha )} =sin(\alpha )$
Рассмотрим левую часть
$\displaystyle \frac{sin(\pi -\alpha )*ctg(\frac{\pi }{2}-\alpha )*cos(2\pi -\alpha )}{tg(\pi +\alpha )*tg(\frac{\pi }{2} +\alpha )*sin(-\alpha )} =\frac{sin(\alpha )*tg(\alpha )*cos(\alpha )}{tg(\alpha )*(-ctg(\alpha ))*-(sin(\alpha ))} =$
$\displaystyle =\frac{sin(\alpha)*\frac{sin(\alpha )}{cos(\alpha )}*cos(\alpha )}{\frac{sin(\alpha )}{cos(\alpha )}*(-\frac{cos(\alpha )}{sin(\alpha )})*(-sin(\alpha ))} =\frac{sin(\alpha )*sin(\alpha )}{-1*(-sin(\alpha ))} =\frac{sin^2(\alpha )}{sin(\alpha )} =sin(\alpha )$
Доказано