Question

What is `cos(pi/12)`?

Answer 1

The answer is: `(sqrt6+sqrt2)/4`

Remembering the formula:

`cos(alpha/2)=+-sqrt((1+cosalpha)/2)`

than, since `pi/12` is an angle of the first quadrant and its cosine is positive so the `+-` becomes `+`,

`cos(pi/12)=sqrt((1+cos(2*(pi)/12))/2)=sqrt((1+cos(pi/6))/2)=`

`=sqrt((1+sqrt3/2)/2)=sqrt((2+sqrt3)/4)=sqrt(2+sqrt3)/2`

And now, remembering the formula of the double radical:

`sqrt(a+-sqrtb)=sqrt((a+sqrt(a^2-b))/2)+-sqrt((a-sqrt(a^2-b))/2)`

useful when `a^2-b` is a square,

`sqrt(2+sqrt3)/2=1/2(sqrt((2+sqrt(4-3))/2)+sqrt((2-sqrt(4-3))/2))=`

`1/2(sqrt(3/2)+sqrt(1/2))=1/2(sqrt3/sqrt2+1/sqrt2)=1/2(sqrt6/2+sqrt2/2)=`

`(sqrt6+sqrt2)/4`