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

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Answer 1 · 2016-03-13T07:48:17

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

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)}#