Question

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1) 4cos 5пи/6 sin 5пи/6 +3tg^2 5пи/6
2) tg 3пи/4 sin3пи/2-ctg пи/2 cos пи/6​

Answer 1

$1)4Cos\frac{5\pi }{6}Sin\frac{5\pi }{6}+3tg^{2}\frac{5\pi }{6}=2Sin\frac{5\pi }{3}+3tg^{2}(\pi+\frac{\pi }{6})= 2Sin(2\pi-\frac{\pi }{3})+3tg^{2}\frac{\pi }{6}=-2Sin\frac{\pi }{3}+3*(\frac{1}{\sqrt{3}})^{2}=-2*\frac{\sqrt{3}}{2}+3*\frac{1}{3}=1-\sqrt{3}$

$2)tg\frac{3\pi }{4}Sin\frac{3\pi }{2}-Ctg\frac{\pi }{2}Cos\frac{\pi }{6}=tg(\pi-\frac{\pi }{4})*(-1)-0*\frac{\sqrt{3}}{2}=-tg\frac{\pi }{4}*(-1)=-1*(-1)=1$