How do you express cos( (5 pi)/6 ) * cos (( pi) /6 ) cos(5π6)cos(π6) 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(5π6)cos(π6)=12[cos(5π6+π6)+cos(5π6π6)]
cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4cos(5π6)cos(π6)=12[cosπ+cos(2π3)]=12(112)=34

Explanation:

cos(x+y)=cos x cos y - sin x sin y" " " " " "cos(x+y)=cosxcosysinxsiny 1st equation
cos(x-y)=cos x cos y+sin x sin y" " " " " "cos(xy)=cosxcosy+sinxsiny 2nd equation

Add first and second equations

cos(x+y)+cos(x-y)=2*cos x cos y+0cos(x+y)+cos(xy)=2cosxcosy+0
cos(x+y)+cos(x-y)=2*cos x cos ycos(x+y)+cos(xy)=2cosxcosy

and then

cos x cos y=1/2[cos(x+y)+cos(x-y)]cosxcosy=12[cos(x+y)+cos(xy)]

Let x=(5pi)/6x=5π6 and y=pi/6y=π6

cos((5pi)/6)*cos(pi/6)=1/2[cos((5pi)/6+pi/6)+cos((5pi)/6-pi/6)]cos(5π6)cos(π6)=12[cos(5π6+π6)+cos(5π6π6)]
cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4cos(5π6)cos(π6)=12[cosπ+cos(2π3)]=12(112)=34

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