Question

В розвязку має бути Π і так далі!

Answer 1

$1)\ \ arccos(-1)+2arccos\dfrac{1}{2}=(\pi -arccos1)+2\cdot \dfrac{\pi}{3}=\pi -0+\dfrac{2\pi }{3}=\dfrac{5\pi}{3}\\\\2)\ \ arccos1-\pi =0-\pi =-\pi \\\\3)\ \ arcsin\Big(-\dfrac{1}{2} \Big)-arccos\dfrac{\sqrt3}{2}=-\dfrac{\pi}{6}-\dfrac{\pi}{6}=-\dfrac{2\pi}{6}=-\dfrac{\pi}{3}\\\\4)\ \ arcsin\dfrac{1}{2}+arccos\Big(-\dfrac{\sqrt3}{2}\Big)=\dfrac{\pi}{6}+\Big(\pi -arccos\dfrac{\sqrt3}{2}\Big)=\dfrac{\pi}{6}+\pi -\dfrac{\pi}{6}=\pi$

$5)\ \ arcsin1+arccos(-1)=\dfrac{\pi}{2}+(\pi -arccos1)=\dfrac{\pi}{2}+\pi -0=\dfrac{3\pi }{2}\\\\6)\ \ arccos\Big(-\dfrac{\sqrt2}{2}\Big)-arccos\dfrac{\sqrt2}{2}=\Big(\pi -arccos\dfrac{\sqrt2}{2}\Big)-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}-\dfrac{\pi }{4}=\dfrac{\pi }{2}$

Answer 2

Объяснение:

$1)\ arccos(-1)+2*arccos\frac{1}{2}= \pi +2*\frac{\pi }{3}=\pi +\frac{2}{3}\pi =1\frac{2}{3}\pi =\frac{5}{3}\pi .\\2)\ arccos1-\pi =0-\pi =-\pi \\3)\ arcsin(-\frac{1}{2})- arccos\frac{\sqrt{3} }{2} = -\frac{\pi }{6} -\frac{\pi }{6}=-\frac{\pi }{3}.\\4)\ arcsin\frac{1}{2} -arccos(-\frac{\sqrt{3} }{2})=\frac{\pi }{6} -\frac{5\pi }{6}=-\frac{4\pi }{6}= -\frac{2}{3} \pi .\\5)\ arcsin1+arccos(-1)=\frac{\pi }{2} +\pi =\frac{3}{2}\pi .\\$

$6)\ arccos(-\frac{\sqrt{2} }{2} )-arccos\frac{\sqrt{2} }{2}=\frac{3\pi }{4} -\frac{\pi }{4} =\frac{2\pi }{4}=\frac{\pi }{2}.$