How do you express #cos( (4 pi)/3 ) * cos (( 7 pi) /4 ) # without using products of trigonometric functions?

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Answer 1 · 2016-10-04T09:33:00

#cos((4pi)/3)cos((7pi)/4)=1/2cos((13pi)/12)+1/2cos((5pi)/12)#

As #cos(A+B)=cosAcosB-sinAsinB# and

#cos(A-B)=cosAcosB+sinAsinB#

#cos(A+B)+cos(A-B)=2cosAcosB#

or #cosAcosB=1/2[cos(A+B)+cos(A-B)]#

Hence #cos((4pi)/3)cos((7pi)/4)#

= #1/2[cos(((4pi)/3)+((7pi)/4))+cos(((4pi)/3)-((7pi)/4))]#

= #1/2[cos((4xx4pi+3xx7pi)/12)+cos((4xx4pi-3xx7pi)/12)]#

= #1/2[cos((16pi+21pi)/12)+cos((16pi-21pi)/12)]#

= #1/2[cos((37pi)/12)+cos((-5pi)/12)]#

= #1/2[cos(2pi+(13pi)/12)+cos((5pi)/12)]#

= #1/2cos((13pi)/12)+1/2cos((5pi)/12)#