How do you evaluate #cos((5pi) / 6) - tan((-7pi) / 4) + sin((13*pi) / 3)#?

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Answer 1 · 2016-10-27T08:14:07

#cos((5pi)/6)-tan(-(7pi)/4)+sin((13pi)/3)#

#=cos(pi-pi/6)+tan(2pi-pi/4)+sin(4pi+pi/3)#

#=-cos(pi/6)-tan(pi/4)+sin(pi/3)#

#=-sqrt3/2-1+sqrt3/2#

#=-1#

How do you evaluate #cos((5*pi) / 6) - tan((-7*pi) / 4) + sin((13*pi) / 3)#?