How do you find the exact value of #cos((2pi)/3)+cos(pi/6)#?

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Answer 1 · 2018-02-24T14:09:43

#(sqrt(3)-1)/2#

The angle #theta = (2pi)/3# is located in the interval #(pi/2,pi)# so that means #cos(theta)=-cos(pi-theta)#.

#cos((2pi)/3) = -cos(pi-(2pi)/3) = -cos(pi/3)#

Now, cosine of #pi/3# (#60°#) is #1/2#.

So #cos((2pi)/3) = -1/2#

#pi/6 = 30°#, so
#cos(pi/6) = sqrt3/2#

In conclusion,

#cos((2pi)/3) + cos(pi/6) = -1/2+sqrt3/2 = (sqrt3-1)/2#.