Question

срочно нужна помощь ​

Answer 1

Объяснение:

1.

a)

$130^0=\frac{130^0*\pi }{180^0} =\frac{13\pi }{18} .\\-210^0=-\frac{210^0*\pi }{180^0}=-\frac{21\pi }{18}=-\frac{7\pi }{6}.$

б)

$-\frac{15\pi }{9} =-\frac{15\pi *180^0}{9*\pi }=15*20^0=300^0.\\ \frac{19\pi }{4} =\frac{19\pi *180^0}{4*\pi }=19*45^0=855^0.$

2.

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

3.

$cos30^0*cos15^0-sin30^0*sin15^0=cos(30^0+15^0)=cos45^0=\frac{\sqrt{2} }{2}.$

4.

$sin\alpha =\frac{4}{5}\ \ \ \ 90^0 < \alpha < 180^0\ \ \ \ sin\frac{\alpha }{2} =?$

$1)\ sin\alpha =2sin\frac{\alpha }{2} *cos\frac{\alpha }{2}=\frac{4}{5}\ |:2\\ sin\frac{\alpha }{2} *cos\frac{\alpha }{2}=\frac{2}{5}\\(sin\frac{\alpha }{2} *cos\frac{\alpha }{2})^2=(\frac{2}{5})^2\\sin^2\frac{\alpha }{2} *cos^2\frac{\alpha }{2}=\frac{4}{25}.$

$2)\ sin^2\alpha +cos^2\alpha =1\\cos^2\alpha =1-sin^2\alpha =1-(\frac{4}{5})^2=1-\frac{16}{25}=\frac{25-16}{25}=\frac{9}{16}.\\ cos\alpha =б\sqrt{\frac{9}{25} }=б\frac{3}{5}\ \ \ \ \ 90^0 < \alpha < 180^0 \ \ \ \ \Rightarrow\\cos\alpha=-\frac{3}{5}.$

$3)\ cos\alpha =cos^2\frac{\alpha }{2} -sin^2\frac{\alpha }{2}=-\frac{3}{5} \\cos^2\frac{\alpha }{2} =sin^2\frac{\alpha }{2} -\frac{3}{5} .$

$4)\ sin^2\frac{\alpha }{2}*cos^2\frac{\alpha }{2} =\frac{4}{25} \\sin^2\frac{\alpha }{2}*(sin^2\frac{\alpha }{2} -\frac{3}{5})=\frac{4}{25} \\sin^4\frac{\alpha }{2}-\frac{3}{5}*sin^2\frac{\alpha }{2}=\frac{4}{25}\ |*25\\25*sin^4\frac{\alpha }{2} -15*sin^2\frac{\alpha }{2} =4.$

Пусть sin²α= t≥0       ⇒

$25t^2-15t-4=0\\D=625\ \ \ \ \sqrt{D}=25\\ t_1=sin^2\frac{\alpha }{2}=-\frac{1}{5} \notin\\ t_2=sin^2\frac{\alpha }{2}=\frac{4}{5}\in\\ sin\frac{\alpha }{2} =б\sqrt{\frac{4}{5} } =б\frac{2}{\sqrt{5} } =б\frac{2\sqrt{5} }{5} \ \ \ \ 90^0 < \alpha < 180^0\ |:2\ \ \ \ 45^0 < \alpha < 90^0\ \ \ \ \Rightarrow\\sin\frac{\alpha }{2} =\frac{2\sqrt{5} }{5} .$

Ответ: sin(α/2)=2√5/5.