Question

How do you find exact value of `cos (pi/12)`?

Answer 1

`cos (pi/12)=sqrt6/4+sqrt2/4`

Explanation:

`cos (pi/12)=cos(pi/4 -pi/6)`
`cos(A-B)=cosAcosB+sinAsinB`
`cos(pi/4 -pi/6)=cos (pi/4)cos(pi/6)+sin(pi/4)sin(pi/6)`
`cos(pi/4 -pi/6)=sqrt2/2*sqrt3/2+sqrt2/2*1/2=sqrt6/4+sqrt2/4`
`cos (pi/12)=sqrt6/4+sqrt2/4`