Question

Convert `5cos((5pi)/3)cos((5pi)/3)` as a sum of trigonometric ratios?

Answer 1

`5cos((5pi)/3)cos((5pi)/3)=5/2(cos((10pi)/3)+cos0)`

= `5/2(cos((10pi)/3)+1)`

Explanation:

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

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

adding the two we get `cos(A+B)+cos(A-B)=2cosAcosB`

Hence `5cos((5pi)/3)cos((5pi)/3)`

`=5/2xx2cos((5pi)/3)cos((5pi)/3)`

`=5/2xx(cos((5pi)/3+(5pi)/3)+cos((5pi)/3-(5pi)/3))`

`=5/2xx(cos((10pi)/3)+cos0)`, but `cos0=1`

Hence `5cos((5pi)/3)cos((5pi)/3)=5/2xx(cos((10pi)/3)+1)`