Question

How to find exact value of `sin(pi/24)`?

Answer 1

`sin(pi/24)=1/2sqrt(2-sqrt(2+sqrt3)`

Explanation:

As `pi/24=180^@/24=(7 1/2)^@`, let us first work out `cos(pi/12)` or `cos15^@`

`cos15^@=cos(45^@-30^@)`

= `cos45^@cos30^@+sin45^@sin30^@`

= `1/sqrt2xxsqrt3/2+1/sqrt2xx1/2`

= `(sqrt3+1)/(2sqrt2)`

Now as `sintheta=sqrt((1-cos2theta)/2)`

`sin(pi/24)=sin(7 1/2)^@`

= `sqrt((1-cos15^@)/2)`

= `sqrt((1-(sqrt3+1)/(2sqrt2))/2)`

= `sqrt((2-(sqrt3+1)/sqrt2)/4)`

= `1/2sqrt(2-(sqrt3+1)/sqrt2)`

= `1/2sqrt(2-(sqrt6+sqrt2)/2)`

= `1/2sqrt(2-sqrt((sqrt6+sqrt2)^2/4)`

= `1/2sqrt(2-sqrt((8+2sqrt12)/4)`

= `1/2sqrt(2-sqrt((8+4sqrt3)/4)`

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