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

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Answer 1 · 2017-03-29T04:58:04

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

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

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)#