Inverse TrigonometryDate: 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/ |
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