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

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

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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)#

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

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