How do you evaluate #sin ((pi)/12)#?

Question

How do you evaluate #sin ((pi)/12)#?

1 Answer

Impact of this question

2413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-23T09:29:50

Using formula #sintheta=sqrt(1/2xx(1-cos(2theta)))#
#sin(pi/12)=sqrt(1/2xx(1-cos(2xxpi/12))#
#=sqrt(1/2xx(1-cos(pi/6))#
#=sqrt(1/2xx(1-sqrt3/2)#
#=sqrt(1/8xx(4-2sqrt3)#
#=sqrt(1/8xx((sqrt3)^2+1^2-2sqrt3*1)#
#=sqrt(1/8xx(sqrt3-1)^2)#
#=1/2xx1/sqrt2xx(sqrt3-1)=(sqrt3-1)/(2sqrt2)#

Science

Math

Humanities

... and beyond

How do you evaluate #sin ((pi)/12)#?

1 Answer

Related questions

Impact of this question