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

1 Answer
Oct 4, 2016

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

Explanation:

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)