How do you find the value of #cot((3pi)/2)#?

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Answer 1 · 2016-02-08T13:30:25

#cot((3pi)/2) = 0#

As you know, #cot(x) = cos(x) / sin(x)#.

Thus, you can compute the value of

#cos((3pi)/2) /sin((3pi)/2)#

instead.

Let's take a look at the #sin# and #cos# functions:

![https://de.wikipedia.org/wiki/Sinus_und_Kosinus]( useruploads.socratic.org )

As you can see, #sin((3pi)/2) = -1# and #cos((3pi)/2) = 0#.

Thus,

#cot((3pi)/2) = 0 / (-1) = 0#.