How do you express cos( (5 pi)/6 ) * cos (( pi) /6 ) without using products of trigonometric functions?

1 Answer

Using Sum:
cos((5pi)/6)*cos(pi/6)=1/2[cos((5pi)/6+pi/6)+cos((5pi)/6-pi/6)]
cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4

Explanation:

cos(x+y)=cos x cos y - sin x sin y" " " " " " 1st equation
cos(x-y)=cos x cos y+sin x sin y" " " " " " 2nd equation

Add first and second equations

cos(x+y)+cos(x-y)=2*cos x cos y+0
cos(x+y)+cos(x-y)=2*cos x cos y

and then

cos x cos y=1/2[cos(x+y)+cos(x-y)]

Let x=(5pi)/6 and y=pi/6

cos((5pi)/6)*cos(pi/6)=1/2[cos((5pi)/6+pi/6)+cos((5pi)/6-pi/6)]
cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4

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