How do you express #cos( (3 pi)/ 2 ) * cos (( 11 pi) /6 ) # without using products of trigonometric functions?

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Answer 1 · 2016-03-05T07:54:14

#cos((3pi)/2)xxcos((11pi)/6)=0#

Here is a graph of #y = cos(x)#
graph{cos(x) [-10, 10, -5, 5]}

#cos(theta)=0# when #theta = pi/2, {3pi}/2, {5pi}/2 ...#

Since #cos((3pi)/2)=0#,

#cos((3pi)/2) xx cos((11pi)/6) = 0 xx cos((11pi)/6)#

#=cos((11pi)/6)0#

Btw

#cos((11pi)/6)=cos(2pi-pi/6)#

#=cos(-pi/6)#
#=cos(pi/6)#
#=sqrt3/2#