What is the value of #\cos (-510)#?

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Answer 1 · 2017-03-17T19:31:08

#- sqrt 3 / 2#

I suppose it's in degrees.

#cos(-a) = cos a = cos (a - 360º)#

#x = cos 510º = cos 150º#

#cos a = - cos (180º - a)#

#x = - cos 30º#

Answer 2 · 2017-03-18T13:51:36

#- sqrt3/2#

Use trig table of special arcs and unit circle -->

#cos (-510) = cos (210 - 2(360)) = cos (210) = #
#= cos (30 + 180) = - cos 30 = - sqrt3/2#