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### Inverse Trigonometry

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Date: 08/22/2001 at 09:21:37
From: Robin
Subject: Inverse Trigonometry

Hello Dr. Maths,

I have just started to do inverse trig.

I found a thing in proving a result. Please tell me about it.

let cot^(-1)(-x)=y--------(1)
=>x=-cot(y)
=>x=cot(-y)
=>cot^(-1)(x)=-y
=>-cot^(-1)(x)=y
=>-cot^(-1)(x)=cot^(-1)(-x)-----using (1)----(2)

Now in the second line we have:

x=-cot(y)
=>x=cot(pi-y)
=>cot^(-1)(x)=pi-y
=>y=pi-cot^(-1)(x)
=>cot^(-1)(-x)=pi-cot^(-1)(-x)-----using(1) -----(3)
from (2) and (3) we have -
pi-cot^(-1)(x)=-cot^(-1)(x)
=>pi=0

but it is not possible. I am not able to find the mistake.

```

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Date: 08/22/2001 at 12:41:58
From: Doctor Peterson
Subject: Re: Inverse Trigonometry

Hi, Robin.

The problem is that the inverse trig functions are inherently
multi-valued functions, for which we traditionally choose a "principal
value" in order to make it a true function. For the cotangent, we
usually use the range 0 < x < pi. (This is similar to taking "the"
square root to be the positive root, avoiding a multi-valued function
by choosing a principal value. Similar problems can arise if you
forget that there are really two square roots.)

See our FAQ for more details:

Trigonometry Formulas - Inverse Trigonometric Functions
http://mathforum.org/dr.math/faq/formulas/faq.trig.html#inverse

Now, when you take the inverse cotangent of both sides, you can't
simply say that

cot^-1(cot(pi-y)) = pi-y

since you don't know that pi-y is in the proper range. For example,
if 0 < x < pi/2, then y < 0 and pi - y > pi is outside the principal
range. To correct for this, you would have to subtract pi to get the
principal value, which will then be -y. This eliminates the apparent

The proper step to take would be

cot^(-1)(x)=pi-y + kpi, for some integer k

If you do this, then you end up with

pi + kpi = 0

which of course is true for k = -1.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Trigonometry

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