How do you express cos( (15 pi)/ 8 ) * cos (( 5 pi) /3 ) without using products of trigonometric functions?

1 Answer
Mar 13, 2016

1/2cos((11pi)/24)+1/2cos((5pi)/24)

Explanation:

As cos(A+B)=cosAcosB-sinAsinB and cos(A-B)=cosAcosB+sinAsinB, adding them gives us

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

Hence cosAcosB=1/2{cos(A+B)+cos(A-B)} and

cos(15pi/8)cos(5pi/3)=1/2{cos((15pi)/8+(5pi)/3)+cos((15pi)/8-(5pi)/3)}

= 1/2{cos((45pi+40pi)/24)+cos((45pi-40pi)/24)}

= 1/2{cos((85pi)/24)+cos((5pi)/24)}

= 1/2{cos((96pi-11pi)/24)+cos((5pi)/24)}

= 1/2{cos(4pi-(11pi)/24)+cos((5pi)/24)}

= 1/2cos((11pi)/24)+1/2cos((5pi)/24)}