How do you evaluate #arccos(sin(13pi/7))#?

Question

4522 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-06T16:33:03

#(-19/14)pi#

Use #sin (pi/2-a)=cos a and arc cos (cos b) =b#.

Here,

#sin ((13/7)pi)=cos (pi/2-(13/7)pi)=cos(-(19/14)pi)#.

The given expression

#= arc cos (cos(-(19/14)pi)#

#=-(19/14)pi=-244.3^o#

Note that, for any arbitrary a, #sin a = cos (pi/2-a)#.

Do not use cos (-b)=cos b, in any intermediate step..

Here, for the start angle #(13/7)pi# (not the principal value), the end

angle is #-(19/14)pi#

Had we used calculator, we would have got the answer as

#114.7^o = (9pi)/14 =(2pi-(19/14)pi)#..