2016年10月5日 · First, $\arccos (\cos x)\ne x$ in your case because $x$ is defined to be from $(\pi, 2\pi)$ and range of $\arccos x$ is $[0,\pi]$, that is $\arccos x$ cannot return ...

2016年10月21日 · From Pythagoras, we have: #sin^2 theta + cos^2 theta = 1# If #x in [-1, 1]# and #theta = arccos(x)# then:. #theta in [0, pi]#

2015年1月24日 · Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1 ...

2014年7月31日 · This is a good definition to memorize, along with #d/dx[arcsin x]# and #d/dx[arctan x]#, since they appear quite frequently in advanced differentiation problems. Answer link Related questions

If $\arccos(x)=y,0\le y\le\pi$ The principal values of $\arcsin$ lies $\in\left[-\dfrac\pi2,\dfrac\pi2\right]$ ...

2020年4月29日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2015年4月26日 · What is $\arccos(x)$? it is the angle given by the ratio of sides of length $x$ and $1$. That is, on a triangle with adjacent side length $x$ and hypotenuse length $1 ...

2020年7月17日 · As I've mentioned, once you differentiate arcsin(x)+arccos(x) and see that this must be a constant, you can just substitute an easy valid x value like x=0 and simplify arcsin(0)+arccos(0) to see that this constant is pi/2.

2016年2月17日 · Following is the proof of the above proposition given in my textbook. Let, $\arccos{x}={\theta}_2$ and $\arcsin{x}={\theta}_1$ $$\sin({\theta}_1 + {\theta}_2)=\sin ...

The same problem happens if you try to expand $\sqrt x$ as a Taylor series. The vertical slope kills you, because polynomials can't do that. The solution is what Alpha does-divide out the "singular part", which in your case is $ \sqrt x.\ \ \frac {\arccos (1-x)}{\sqrt x}$ is nicely behaved at the origin and can give you a Taylor series.