How do you find the 6 trigonometric functions for 600 degrees?

2 Answers

#sin(600^"o") = -sqrt(3)/2#
#cos(600^"o") = -1/2#
#sec(600^"o") = -2#
#csc(600^"o") = -(2sqrt(3))/3#
#tan(600^"o") = sqrt3#
#cot(600^"o") = sqrt(3)/3#

Explanation:

First thing, let's see how many full loops we can get out of that angle, #theta = 600^"o" = 360^"o" + 240^"o" = 2*360^"o" - 120^"o"#

Why this is important, you may ask? It's because every 360^"o" degrees we have a full loop and return to where we were. So for the two main trig functions, #sin(x)# and #cos(x)# we can say that

#sin(600^"o") = sin(240^"o") = sin(-120^"o")#
#cos(600^"o") = cos(240^"o") = cos(-120^"o")#

And since every other function is a ratio of one or two of these functions, we can work with that. Now, we know that
#sin(-theta) = -sin(theta)# and #cos(-theta) = cos(theta)#

So we can say that
#sin(600^"o") = -sin(120^"o")#
#cos(600^"o") = cos(120^"o")#

We can then use the angle sum / double angle formula, since #120^"o" = 60^"o"+60^"o"#
#-sin(120^"o") = -(sin(60^"o")cos(60^"o") + sin(60^"o")cos(60^"o"))#
#cos(120^"o") = cos(60^"o")cos(60^"o") - sin(60^"o")sin(60^"o"))#

Since #60^"o"# is a special angle we know that #sin(60^"o") = sqrt(3)/2# and #cos(60^"o") = 1/2#, so

#-sin(120^"o") = -(sqrt(3)/2*1/2 + sqrt(3)/2*1/2) = -sqrt(3)/2#
#cos(120^"o") = 1/2*1/2- sqrt(3)/2*sqrt(3)/2 = -2/4 = -1/2#

And from that, we just evaluate the others by division:
#sec(600^"o") = 1/cos(600^"o") = -1/(1/2) =-2#
#csc(600^"o") = 1/sin(600^"o") = -1/(sqrt(3)/2) = -2/sqrt(3) = -(2sqrt(3))/3#
#tan(600^"o") = sin(600^"o")/cos(600^"o") = (sqrt(3)/2)/(1/2) = sqrt3#
#cot(600^"o") = 1/tan(600^"o") = 1/(sqrt(3)) = sqrt(3)/3#

Sep 23, 2015

Find the 6 trig functions of 600 deg

Explanation:

600 = 240 + 360 deg
#sin 600 = sin 240 = - sin 60 = - sqrt3/2#
#cos 600 = cos 240 = - cos 60 = - 1/2#
#tan 600 = tan 240 = (-sqrt3/2)/-(1/2) = sqrt3# (arc 240 is in Quadrant III, its tan is positive)
#cot 600 = 1/(tan) = 1/sqrt3 = sqrt3/3#
#sec 600 = 1/(cos) = - 2#
#csc 600 = 1/(sin) = - (2sqrt3)/3#