How do you find the six trigonometric functions of #-3pi# degrees?

1 Answer
Jun 18, 2018

#cos(-3pi) = -1#
#sin(-3pi) = 0#
#tan(-3pi)=0#
#sec(-3pi) = -1#
#csc(-3pi)# is undefined
#cot(-3pi)# is undefined

Explanation:

Since the whole #360°# angle translates to #2pi# radians, an angle of #pi# randians means half a turn around the circumference. So, #-3pi# means that you two three half turns clockwise.

Actually, you may notice that doing a half turn clockwise or counter clockwise has the same result: you will end up in the opposite point of where you started any way.

After two half turns we are where we started, so performing three half turn is the same as performing only one half turn, i.e. #pi# degree.

So, we cleared that #-3pi# radians intercepts the same point as #pi# radians.

Since we start from the point #(1,0)#, after half turn we are at the point #(-1,0)#. And since any point on the unit circle has coordinates #(cos(alpha),sin(alpha))#, we have

#cos(-3pi) = cos(pi) = -1#
#sin(-3pi) = sin(pi) = 0#

We can immediately deduce the remaining functions:

#tan(-3pi)=tan(pi) = sin(pi)/cos(pi) = 0/(-1)=0#

The other are simply the inverse of those we just wrote:

#sec(-3pi) = sec(pi) = 1/cos(pi) = 1/-1 = -1#
#csc(-3pi) = csc(pi) = 1/sin(pi) = 1/0 \to# undefined
#cot(-3pi) = cot(pi) = 1/tan(pi) = 1/0 \to# undefined