Question

Вычислите.
Задание на фото.

Answer 1

$1)Cos\frac{\pi }{6}+Sin(-\frac{\pi }{3})-2Ctg(-\frac{\pi }{4})+2Cos\pi=Cos\frac{\pi }{6}-Sin\frac{\pi }{3}+2Ctg\frac{\pi }{4}+2Cos\pi =\\\\=\frac{\sqrt{3} }{2}-\frac{\sqrt{3} }{2}+2*1+2*(-1)=2-2=\boxed0\\\\\\2)2Sin\frac{\pi }{6}+tg(-\frac{\pi }{4})+2Cos(-\frac{\pi }{6})+Ctg(-\frac{\pi }{6})=\\\\=2Sin\frac{\pi }{6}-tg\frac{\pi }{4}+2Cos\frac{\pi }{6}-Ctg\frac{\pi }{6}=2*\frac{1}{2}-1+2*\frac{\sqrt{3}}{2}-\sqrt{3}=1-1+\sqrt{3}-\sqrt{3} =\boxed0$

Answer 2

а)

$\cos\dfrac{\pi}{6}+\sin\left(-\dfrac{\pi}{4}\right)-2\cdot\text{ctg}\left(-\dfrac{\pi}{4}\right)}+2\cdot\cos\pi=\dfrac{\sqrt{3} }{2}-\dfrac{\sqrt{3} }{2}+2\cdot 1+2\cdot (-1)=2-2=0$

б)

$2\cdot\sin\dfrac{\pi}{6}+\text {tg}\left(-\dfrac{\pi}{4}\right)+2\cdot\cos\left(-\dfrac{\pi}{6}\right)+\text {ctg}\left(-\dfrac{\pi}{6}\right)=2\cdot\dfrac 1 2-1+2\cdot\dfrac{\sqrt{3} }{2}-\sqrt 3=1-1+\sqrt 3-\sqrt 3=0$