2016年9月2日 · Range: [-pi/2,pi/2] Domain: [-1,1] The following is a fragment from my lecture about y=arcsin x presented on UNIZOR.COM. If you go to this very useful Web site, click Trigonometry - Inverse Trigonometric Functions - y=arcsin(x). The original sine function defined for any real argument does not have an inverse function because it does not establish a one-to …

2015年12月31日 · We could choose another range for each inverse trigonometric function. For example, we can pick $[0,\pi]$ to be the range of $\sin^{-1}x$. EDIT. I've understood why the range of the sine and cosine has to be $[-\pi/2,\pi/2]$ and $[0,\pi]$ respectively. I'm still wondering why can't we define the range of the tangent as $[0,\pi]$

2017年7月13日 · The range for the function is the same as the range for the cosine function, -1<=f(x)<=1 Use the identity cos(x) = +-sqrt(1-sin^2(x)) f(x) = +-sqrt(1-sin^2(sin^-1(x))) The sine and the inverse sine function cancel; leaving us with x^2: f(x) = +-sqrt(1-x^2); -1<=x<=1 We cannot know whether we use the + or the - sign without more information ...

$\begingroup$ Again my argument is that if θ1=5π/4 then θ2=−π/4, which is out of our desired range of [0,2π]. $\endgroup$ – kaka Commented Nov 21, 2014 at 15:48

2018年10月7日 · $\begingroup$ No. They have the same domain, which is $[-1,1]$. They just end up in different ranges. So, you can add $\arcsin x$ and $\arccos x$ however you want.

2015年11月23日 · Domain: [-1,+1] Range: [-pi/2,+pi/2] The argument for arcsin(x) (i.e. x) must be a value which can be returned by the sin function; since the sin function has a range of [-1,+1] arcsin has a domain of [-1,+1] The range of the arcsin function is a matter of definition. In order for arcsin to be a function its range is defined to be [-pi/2,+pi/2]; without this defined restriction …

The logarithmic form of $\text{arcsin}(x)$ and its implications Hot Network Questions Predatory Journal keeps sending me Emails inviting me to publish in their journal

2018年12月4日 · $$\arcsin(\sin(x)) = \sin(\arcsin(x)) = x $$ Now to address your other question. $\arcsin(x)$ IS a function. If you wish to know it's domain and range, first consider the domain and range of $\sin(x)$ This is a bounded periodic function whose domain is all of the real numbers and whose range is from -1 to 1 inclusive.

2016年5月31日 · The definition of a function says you can get from any point in the domain to a unique point in the range; it says nothing about going from the range to the domain. As a real-life analogy, there are machines that can turn standing trees into wood chips, but not (yet) any machine that can turn wood chips into a standing tree.

The range of $\sin x$ is $[-1,1]$, which is the value you will try to take the $\arcsin$ of. The $\arcsin$ is happy doing that, so the domain of $\arcsin(\sin(x))$ is all the real line. The compound function will not be the identity over the whole line, but …