Question

Вычилслите а) arccos v2/2 + 2 arccos(-1/2)-arccos 0 б) arccos(ctg3п/4) в) sin(arccos(=v3/2))

Answer 1

1) $arccos( \frac{ \sqrt{2}}{2})+2arccos(- \frac{1}{2})-arccos0=$$\frac{ \pi }{4}+2*( \pi - \frac{ \pi }{3})-\frac{ \pi }{2}=\frac{ \pi }{4}+\frac{4 \pi }{3}-\frac{ \pi }{2}=\frac{ 3\pi+16 \pi -6 \pi }{12}=\frac{13 \pi }{12}$

2) $arccos(ctg( \frac{3 \pi }{4}))=arccos(ctg( \pi -\frac{\pi }{4}))=arccos(-ctg(\frac{\pi }{4}))=$$\pi -arccos(ctg(\frac{\pi }{4}))= \pi -arccos(1)= \pi -0= \pi$

3) $sin(arccos(\frac{\sqrt{3}}{2}))=sin(\frac{ \pi }{6})=\frac{1}{2}$