How do I find the exact value of #cos ((5pi)/4)+cot((5pi)/3)#?

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Answer 1 · 2017-11-22T10:15:55

See explanation.

First I calculate both trigonometric functions:

#cos((5pi)/4)=cos(pi+pi/4)=-cos(pi/4)=-sqrt(2)/2#

#cot((5pi)/3)=cot(2pi-pi/3)=cot(-pi/3)=-cot(pi/3)=#

#=-sqrt(3)/3#

Now if we add voth expressions we get:

#-(sqrt(2)/2+sqrt(3)/3)=-((3sqrt(2))/6+(2sqrt(3))/6)=#

#=(-3sqrt(2)-2sqrt(3))/6#