How do you express #cos( (3 pi)/ 2 ) * cos (( 3 pi) /4 ) # without using products of trigonometric functions?

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Answer 1 · 2016-05-03T09:54:54

#cos((3pi)/2)*cos((3pi)/4)=0#

As #cos(A-B)=cosAcosB-sinAsinB# and #cos(A+B)=cosAcosB+sinAsinB#, adding them

#2cosAcosB=cos(A+B)+cos(A-B)#

or #cosAcosB=1/2xxcos(A+B)+1/2xxcos(A-B)#

Hence #cos((3pi)/2)*cos((3pi)/4)#

= #1/2xxcos(((3pi)/2)+((3pi)/4))+1/2xxcos(((3pi)/2)-((3pi)/4))#

= #1/2{cos((9pi)/4)+cos((3pi)/4)}#

= #1/2{cos(2pi+pi/4)+cos(pi-pi/4)}#

= #1/2{cos(pi/4)-cos(pi/4)}=0#