Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]sin^2120'+cos^2150'=[sin(180'-60')]^2+[cos(180'-30')]^2=\\=(sin60')^2+(-cos30')^2=(\frac{\sqrt{3} }{2} )^2+(-\frac{\sqrt{3} }{2})^2=\frac{3}{4} +\frac{3}{4}=\frac{6}{4} =1,5[/tex]
[tex]\frac{cos120'}{sin^2330} +\frac{tg150'}{tg210'}=\frac{cos(180'-60')}{[sin(360'-30')]^2} +\frac{tg(180'-30')}{tg(180'+30')} =\\\frac{-cos60'}{(-sin30')^2} +\frac{-tg30'}{tg30'} =-\frac{1}{2} :(-\frac{1}{2})^2+(-\frac{\sqrt{3} }{3}):(\frac{\sqrt{3} }{3} ) =-\frac{1}{2}*\frac{4}{1}+(-\frac{\sqrt{3} }{3})*\frac{3}{\sqrt{3} }}=-2-1=-3[/tex]
[tex]cos330'+tg120'*tg330'=cos(360'-30')+tg(180'-60')*tg(270'+60')=\\=cos30'+(-tg60')*(-ctg60')=\frac{\sqrt{3} }{2} +\sqrt{3} *\frac{\sqrt{3} }{3} =1+\frac{1}{2} \sqrt{3}[/tex]
[tex]\frac{cos^2150'*tg210'*tg135'}{sin^2150'+cos^2210'} =\frac{[cos(180'-30')]^2*tg(270'-60')*tg(180'-45')}{[sin(180'-30')]^2+[cos(180'+30')]^2} =\\=\frac{-cos30'*ctg60'*(-tg45')}{(sin30')^2+(-cos30')^2}=[-\frac{\sqrt{3} }{2}*\frac{\sqrt{3} }{3} *(-1)]:[(\frac{1 }{2} )^2+(-\frac{\sqrt{3} }{2} )^2]=\\=\frac{1}{2} :1=\frac{1}{2}[/tex]
