Question

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

Answer 1

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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