Normal And Inverse Trig Functions Cancel Each Other Out

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Normal and inverse trig functions cancel each other out
  • Context: High School
  • Thread starter Thread starter TSN79
  • Start date Start date Nov 9, 2004
  • Tags Tags Functions Inverse Normal Trig Trig functions
Click For Summary

Discussion Overview

The discussion centers around the properties of normal and inverse trigonometric functions, particularly focusing on the equations involving sine and arcsine, and tangent and arctangent. Participants explore the conditions under which these functions can be considered inverses, including the implications of function injectivity and the effects of domain restrictions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that while \(\sin(\arcsin(x)) = x\) holds true, \(\arcsin(\sin(x)) = x\) is not universally valid due to the non-injectivity of the sine function outside specific intervals.
  • One participant points out that \(\arcsin(x)\) is undefined for \(|x| > 1\), which affects the validity of the equations.
  • Another participant discusses the necessity of restricting the domain of sine to \([- \frac{\pi}{2}, \frac{\pi}{2}]\) to ensure that \(\arcsin(\sin(x)) = x\) holds true.
  • There is a mention of the tangent function's similar issues with injectivity, and that \(\arctan(\tan(x)) = x\) is valid only when \(x\) is restricted to \((- \frac{\pi}{2}, \frac{\pi}{2})\).
  • One participant raises a question about finding solutions for \(\tan(x/2) = 0\) and whether using \(\arctan(\tan(x/2)) = x/2\) would work under certain conditions.
  • Another participant clarifies that the period of \(\tan(x/2)\) is \(2\pi\) and discusses the implications of this on the range of the inverse function.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the inverse relationships for trigonometric functions, particularly regarding domain restrictions and injectivity. There is no consensus on the conditions under which these relationships hold true, and multiple competing views remain throughout the discussion.

Contextual Notes

Limitations include the need for specific domain restrictions for the sine and tangent functions to ensure their inverses are valid. The discussion also highlights the undefined nature of arcsine for inputs outside the range of \([-1, 1]\) and the implications for the validity of certain equations.

TSN79 Messages 422 Reaction score 0 I'm a bit confused at the moment. All my books say that normal and inverse trig.functions cancel each other out like this \sin(\arcsin(x))=x and \arcsin(sin(x))=x But when I try this out on my calculator - TI-89 - it only wants to recognize the first equation as being equal to x. Is that true or...? :confused: Last edited: Nov 9, 2004 Mathematics news on Phys.org
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Galileo Science Advisor Homework Helper Messages 1,980 Reaction score 7 For a function to have an inverse, it needs to be injective and surjective. The problem is that sin(x) is generally not injective, however it is injective on (eg) the interval [-\pi/2,\pi/2]. This interval can be chosen as the domain of sin(x) and becomes the range of arcsin(x), so for x in this interval the second equation will hold as well. Last edited: Nov 10, 2004 TSN79 Messages 422 Reaction score 0 But since my calculator always accepts sin(arcsin(x)) as being equal to x, does that mean that this equation is always true no matter what? :confused: Galileo Science Advisor Homework Helper Messages 1,980 Reaction score 7 That's funky. You cannot take the arcsin of x if |x|&gt;1, since such an x wouldn't be in the range of sin(x). My calculator gives an error, as it should. (Unless you use complex numbers). krab Science Advisor Messages 892 Reaction score 3 Think it through. Try x=360 degrees. sin(360 deg.)=0, but arcsin(0)=0. So arcsin(sin(x)) is not equal to x. OTOH, sin(arcsin(x))=x HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Another function that does not have a "true" inverse is f(x)= x2 and you might find it easier to look at that function (you can do the calculations yourself rather than use a calculator). If we restrict the domain to non-negative numbers, THEN we can use f-1(x)= &radic;(x) as the inverse. &radic;(x) only makes sense when x is positive. For any positive number, if you first take the square root and then square, you get x back again: f(f-1(x))= x. In particular, if x= 4, f-1(4)= &radic;(4)= 2 and f(2)= 22= 4 again. However, f(x) is defined for x positive or negative. If x= 2, f(2)= 22= 4 and then f-1(4)= &radic;(4)= 2. In that case, f-1[\sup](f(x))= x. But if x= -2, f(-2)= (-2)2= 4 and then f-1(4)= &radic;(4)= 2. In that case, f-1(f(x)) is NOT equal to x. tmwong Messages 24 Reaction score 0 sin(arcsin(x))=x true while arcsin(sin(x))=x is not true. Galileo Science Advisor Homework Helper Messages 1,980 Reaction score 7 sin(arcsin(x))=x does not make sense if |x|>1, since it is undefined. It's only true if |x|<=1. TSN79 Messages 422 Reaction score 0 So how would I go about it to find the solutions for let's say tan(x/2)=0 ? I thought I could make it easier by using arctan(tan(x/2))=x/2, but you're telling me that this wouldn't work? What if I decided to look only at a part of tan(x/2) that is increasing or decreasing? Let's say from -pi to pi. Would this work? Because that's the requirement for a trig.function to have an inverse isn't it? Would arctan(tan(x/2))=x/2 be true then? Galileo Science Advisor Homework Helper Messages 1,980 Reaction score 7 tan(x) has the same problem: It isn't injective. However, it IS injective if it is restricted to the interval (-\frac{\pi}{2},\frac{\pi}{2}) Then for x in this interval \arctan(\tan(x))=x holds. (Notice that range of arctan(x) is (-\frac{\pi}{2},\frac{\pi}{2})) If you wish to solve \tan(x/2)=0, then using arctan will give you an answer that lies in the interval (-\frac{\pi}{2},\frac{\pi}{2}). (In this case x=0). For a complete answer, you need additional info. In this case you can use that the tangent is periodic (period \pi) so that x/2=k\pi where k is an integer will give the general solution). Last edited: Nov 10, 2004 TSN79 Messages 422 Reaction score 0 Wouldn't period of tan(x/2) be 2pi? When the interval of tan(x) is restricted to <-pi/2 , pi/2>, the range of arctan(x) is the same. But when the interval of tan(x/2) is <-pi , pi>, but now the range of arctan(x/2) is not the same. Why is this? Shouldn't the range of arctan(x/2) the same as the interval of tan(x/2)? Galileo Science Advisor Homework Helper Messages 1,980 Reaction score 7 Yes, the period of tan(x/2) is 2\pi (I meant the period of tan(x) in my post). If y=\tan(x/2). Then for x \in (-\pi,\pi): \arctan(y)=x/2 x=2\arctan(y) So the inverse function of tan(x/2) is 2arctan(x) and has range (-\pi,\pi)

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