Question

Find the exact value of `sin(pi/12)` and `cos(pi/12)`?

Answer 1

`sin(pi/12)=(sqrt3-1)/(2sqrt2)`

`cos(pi/12)=(sqrt3+1)/(2sqrt2)`

Explanation:

As `cos2A=2cos^2A-1`

`cos(2xxpi/12)=2cos^2(pi/12)-1`

or `cos(pi/6)=sqrt3/2=2cos^2(pi/12)-1`

or `2cos^2(pi/12)-1-sqrt3/2=0`

or `2cos^2(pi/12)-(2+sqrt3)/2=0`

or `cos^2(pi/12)=(2+sqrt3)/4`

and `cos(pi/12)=sqrt(2+sqrt3)/2=sqrt(4+2sqrt3)/(2sqrt2)`

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

= `sqrt((sqrt3)^2+1^2+2xxsqrt3xx1)/(2sqrt2)`

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

Further `cos2A=1-2sin^2A`

`cos(pi/6)=1-2sin^2(pi/12)`

or `sqrt3/2=1-2sin^2(pi/12)`

or `2sin^2(pi/12)=1-sqrt3/2`

or `sin^2(pi/12)=(2-sqrt3)/4`

or `sin(pi/12)=(sqrt(2-sqrt3)/2`

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

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

= `sqrt((sqrt3)^2+1^2-2xxsqrt3xx1)/(2sqrt2)`

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