Question

How do you find the value of `arccos(-sqrt2/2)`?

Answer 1

The value is `(3pi)/4`

Explanation:

Given `cos^-1(-sqrt2/2)=pi-cos^-1(sqrt2/2)(sqrt2/sqrt2)`
`rArrpi-cos^-1(1/sqrt2)rArrpi-pi/4rArr(3pi)/4`