Question

помогите пожалуйста с алгеброй

Answer 1

Объяснение:

1.

$\frac{cos68^0-cos22^0}{sin68^0-sin22^0} =\frac{-2*sin\frac{68^0+22^0}{2}*sin\frac{68^0-22^0}{2} }{2*sin\frac{68^0-22^0}{2}*cos\frac{68^0+22^0}{2}}=-\frac{sin45^0*sin23^0}{sin23^0*cos45^0} =-\frac{\frac{\sqrt{2} }{2} }{\frac{\sqrt{2} }{2} }=-1.$

2.

$a)\ (\frac{sin\alpha }{1+cos\alpha }-\frac{sin\alpha }{1-cos\alpha })*\frac{1}{2}tg(-\alpha )=-sin\alpha * (\frac{1 }{1+cos\alpha }-\frac{1 }{1-cos\alpha })*\frac{1}{2}tg\alpha=$

$=-sin\alpha *\frac{1-cos\alpha -1-cos\alpha }{(1+cos\alpha )*(1-cos\alpha )} *\frac{1}{2} tg\alpha =\frac{2*sin\alpha*cos\alpha }{1-cos^2\alpha }*\frac{1}{2}*tg\alpha =\frac{sin\alpha *cos\alpha *sin\alpha }{sin^2\alpha *cos\alpha }=1.$

$b)\ (1-cos2\alpha )*tg(\frac{\pi }{2} +\alpha )=(sin^2\alpha +cos^2\alpha -cos^2\alpha+sin^2\alpha )*(-ctg\alpha )=$

$=-2*sin^2\alpha *\frac{cos\alpha }{sin\alpha } =-2*sin\alpha *cos\alpha =-sin2\alpha .$

3.

$cos\alpha =-\frac{4}{5} \ \ \ \ \ 180^0 < \alpha < 270^0\ \ \ \ \ cos\frac{\alpha }{2}=?\\ cos\alpha =cos^2\frac{\alpha }{2}-sin^2\frac{\alpha }{2}=-\frac{4}{5}\\ cos^2\frac{\alpha }{2} =sin^2\frac{\alpha }{2}-\frac{4}{5} \\ cos^2\frac{\alpha }{2}+cos^2\frac{\alpha }{2} =sin^2\frac{\alpha }{2}+cos^2\frac{\alpha }{2}-\frac{4}{5} \\ 2*cos^2\frac{\alpha }{2}=1-\frac{4}{5} \\ 2*cos^2\frac{\alpha }{2}=\frac{1}{5}\ |:2\\cos^2\frac{\alpha }{2}=\frac{1}{10}$

$cos\frac{\alpha }{2}=б\sqrt{\frac{1}{10} } =б\frac{\sqrt{10} }{10} } \ \ \ \ \ 180^0 < \alpha\ < 270^0\ |:2\ \ \ \ \ 90^0 < \frac{\alpha }{2} < 135^0\ \ \ \ \Rightarrow\\cos\frac{\alpha }{2} =-\frac{\sqrt{10} }{10}.$