Question

Помогите пожалуйста ДАЮ 60 БАЛЛОВ

Answer 1

Ответ:

Объяснение:

1.

$a)7*sin0^{0} +5*cos60^{0} =0+5*\frac{1}{2} =\frac{5}{2} =2,5.\\b)4*sin\frac{\pi }{2} -2*tg\frac{\pi }{4} =4*1-2*1=4-2=2.$

2.

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

3.

$sin\alpha =-\frac{12}{13} ;\pi <\alpha <\frac{3\pi }{2} ;cos\alpha= ?;tg\alpha =?\\cos\alpha =\sqrt{1-sin^{2}\alpha } =\sqrt{1-(-\frac{12}{13})^{2} }=\sqrt{1-\frac{144}{169} }=\sqrt{\frac{25}{169} } =б\frac{5}{13} .\\\pi <\alpha <\frac{3\pi }{2} \\cos\alpha =-\frac{5}{13} .\\ tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{-\frac{12}{13} }{-\frac{5}{13} } \\tg\alpha =\frac{12}{5}.$

4.

$[( \frac{1+cos\alpha }{sin\alpha } )^{2}+1 ]:\frac{1+cos\alpha }{sin^{2} \alpha } =[{\frac{(1+cos\alpha )^{2} }{sin^{2} \alpha }+1]*\frac{sin^{2} \alpha }{1+cos\alpha } =\frac{1+2*cos\alpha+cos^{2}\alpha+sin^{2}\alpha }{sin^{2}\alpha }*\frac{sin^{2}\alpha }{1+cos\alpha } =$$=\frac{1+2*cos\alpha +1}{sin^{2}\alpha } *\frac{sin^{2} \alpha }{1+cos\alpha } =\frac{2+2*cos\alpha }{sin^{2}\alpha } *\frac{sin^{2} \alpha }{1+cos\alpha } =\frac{2*(1+cos\alpha)*sin^{2} \alpha }{sin^{2}\alpha *(1+cos\alpha) } =2.$