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

We want to find replacement angles for `pi/12" that will produce exact values "`

These must come from : `pi/6 , pi/3 , pi/4`

`rArr sin(pi/12) = sin(pi/3 - pi/4 )`

Using the appropriate `color(blue)" Addition formula "`

`color(red)(|bar(ul(color(white)(a/a)color(black)( sin(A ± B) = sinAcosB ± cosAsinB )color(white)(a/a)|)))`

`rArr sin(pi/3 - pi/4) = sin(pi/3)cos(pi/4) - cos(pi/3)sin(pi/4)`

Extract `color(blue)" exact values from triangles "`
`sin(pi/3) = (sqrt3)/2 , sin(pi/4) = 1/(sqrt2)`
and `cos(pi/3) = 1/2 , cos(pi/4) = 1/(sqrt2)`
now substitute into the right side of the expansion.

`= (sqrt3)/2xx1/(sqrt2) - 1/2xx1/(sqrt2) = (sqrt3)/(2sqrt2)-1/(2sqrt2)`

`= (sqrt3 - 1)/(2sqrt2) " and rationalising the denominator "`

gives `((sqrt3 - 1)xxsqrt2)/(2sqrt2xxsqrt2)= (sqrt6 - sqrt2)/4`

Apply the trig identity:
`cos 2a = 1 - 2sin^2 a`
`cos ((2pi)/12) = cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)`
`2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2`
`sin^2 (pi/12) = (2 - sqrt3)/4`
`sin (pi/12) = +- sqrt(2 - sqrt3)/2`
Since `sin (pi/12)` is positive, then only the positive answer is accepted.
`sin (pi/12) = sqrt(2 - sqrt3)/2`

Check by calculator.
Calculator --> `sin (pi/12) = sin 15^@ = 0.26`.
`sqrt(2 - sqrt3)/2 = sqrt(0.27)/2 = 0.52/2 = 0.26`. OK.