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How do you evaluate #3cos((17pi)/6)+2cos((-5pi)/3 )#?

1 Answer

A. S. Adikesavan

Dec 1, 2016

#(-3sqrt3+2)/2=-1.2132, #nearly.

Explanation:

Use #cos (-a) = cos a# and cosine is negative in #Q_2# and positive in #Q_4#.

#3 cos (17/6pi)+ 2 cos (-5/3pi)#

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

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

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

#=(-3sqrt3+2)/2#

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