Convert 5cos((5pi)/3)cos((5pi)/3)5cos(5π3)cos(5π3) as a sum of trigonometric ratios?

1 Answer
Mar 29, 2017

5cos((5pi)/3)cos((5pi)/3)=5/2(cos((10pi)/3)+cos0)5cos(5π3)cos(5π3)=52(cos(10π3)+cos0)

= 5/2(cos((10pi)/3)+1)52(cos(10π3)+1)

Explanation:

As cos(A+B)=cosAcosB-sinAsinBcos(A+B)=cosAcosBsinAsinB and

cos(A-B)=cosAcosB+sinAsinBcos(AB)=cosAcosB+sinAsinB

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

Hence 5cos((5pi)/3)cos((5pi)/3)5cos(5π3)cos(5π3)

=5/2xx2cos((5pi)/3)cos((5pi)/3)=52×2cos(5π3)cos(5π3)

=5/2xx(cos((5pi)/3+(5pi)/3)+cos((5pi)/3-(5pi)/3))=52×(cos(5π3+5π3)+cos(5π35π3))

=5/2xx(cos((10pi)/3)+cos0)=52×(cos(10π3)+cos0), but cos0=1cos0=1

Hence 5cos((5pi)/3)cos((5pi)/3)=5/2xx(cos((10pi)/3)+1)5cos(5π3)cos(5π3)=52×(cos(10π3)+1)