Inverse Trigonometry

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.

Please help me. Thanks.

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 
contradiction.

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/