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 $cos((3pi)/2)*cos((3pi)/4)$

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

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

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

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

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