How do you find exact value of #cos (pi/12)#?

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How do you find exact value of #cos (pi/12)#?

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Answer 1 · 2016-03-30T06:09:30

#cos (pi/12)=sqrt6/4+sqrt2/4#

Explanation:

#cos (pi/12)=cos(pi/4 -pi/6)#
#cos(A-B)=cosAcosB+sinAsinB#
#cos(pi/4 -pi/6)=cos (pi/4)cos(pi/6)+sin(pi/4)sin(pi/6)#
#cos(pi/4 -pi/6)=sqrt2/2*sqrt3/2+sqrt2/2*1/2=sqrt6/4+sqrt2/4#
#cos (pi/12)=sqrt6/4+sqrt2/4#

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How do you find exact value of #cos (pi/12)#?

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