How do you find the value of #cos ((23pi)/12)# using the double or half angle formula?

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Answer 1 · 2016-06-24T13:09:29

#cos((23pi)/12)=sqrt(2+sqrt3)/2#

#cos((23pi)/12)=cos(2pi-pi/12)=cos(pi/12)#

Using formula #cos2A=2cos^2A-1#, if #A=pi/12#

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

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

#2cos^2(pi/12)=sqrt3/2+1=(2+sqrt3)/2# or

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

#cos(pi/12)=sqrt(2+sqrt3)/2#

Hence #cos((23pi)/12)=sqrt(2+sqrt3)/2#