Question

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

Answer 1

`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)}`