Arccot(tan2pi/3)?

1 Answer
Dean R.
May 3, 2018

# text{Arc}text{cot}(tan({2pi}/3))#

# = text{Arc}text{cot}(cot(pi/2 - {2pi}/3))#

# = text{Arc}text{cot}(cot(-pi/6))#

# = {5pi}/6 #

Explanation:

We're given a tangent of an angle and asked for the inverse cotangent. I'll take the capital A Arccot as asking for the principal value.

The "co" in cotangent and cosine and cosecant means "complementary." It refers to complementary angles. It's reminding us of the identity, e.g.,

# cot theta = tan(pi/2 - theta)#

That's all we need here:

# text{Arc}text{cot}(tan({2pi}/3))#

# = text{Arc}text{cot}(cot(pi/2 - {2pi}/3))#

# = text{Arc}text{cot}(cot(-pi/6))#

# = -pi/6 + pi #

# = {5pi}/6 #

I remember that cotangent is an interesting choice for triangle angles because it's uniquely defined for #0 < theta < pi,# angles of a real triangle, and undefined at #0# and #pi#, the degenerate triangle. So the principal value of the inverse cotangent is the first two quadrants, so we need our answer from the second quadrant, so tempting as it is we don't just write #-pi/6# from the fourth quadrant.

When we're faced with the slightly different

#theta = text{arccot}(cot (-pi/6)) #

where the small letter indicates the multivalued inverse cotangent, all the values that have a cotangent equal to #cot(-pi/6)#, we can write

#theta = -pi/6 + k pi quad# for integer #k#

Because this is all the values, we don't have to be concerned that the one we wrote down, #-pi/6# isn't the principal value, as long as the principal value comes up for some integer value of #k#.

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