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

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Answer 1 · 2016-02-12T23:41:27

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

$cos(x+y)=cos x cos y - sin x sin y" " " " " "$ 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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How do you express cos( (5 pi)/6 ) * cos (( pi) /6 ) cos( (5 pi)/6 ) * cos (( pi) /6 ) without using products of trigonometric functions?