2020年4月11日 · Whenever I calculate inverse tangent on the calculator it only gives the correct angle when the point (X, Y) is in the first and fourth quadrant. Whenever the point lies in the second or third quad...

2022年1月23日 · By definition of the inverse tangent, $\tan^{-1}y$ is some angle between $-90^\circ$ and $90^\circ$, but it's not the only one that satisfies $\tan x=y$. To get other solutions, you have to add or subtract a multiple of $180^\circ$. As an …

2016年1月27日 · Thank you Ng Chung Tak for an excellent answer.I am trying to answer the Stack Exchange Mathematics Question, "Help with the indefinite integral $\int \frac{dx}{2x^4 + 3x^2 + 5}$ ", and I needed to put a few more steps into the answer to make it clear how you go from line to line.

2021年1月11日 · $\begingroup$ (+1) One other reason I dislike the $\tan^{-1}$ notation is because it suggests that the $\tan$ function has an inverse. This is not true: $\tan^{-1}(\tan(x)) \neq x$ in general. People often say '$\tan$ has an inverse on a restricted domain', but when you restrict the domain of a function, you are creating a new one.

2020年3月18日 · I see here a list of inverse trigonometric functions written in terms of logarithms. The ones I'm most interested in for the purposes of this question are $\arcsin{z}=-i\ln\left(iz+\sqrt{1-z^2}\ri...

2015年9月2日 · $\tan^{-1}$ denotes the inverse tangent function, AKA the arc tangent (the angle the tangent of which is the given number). When applied to an argument, you spell $$\tan^{-1}(x)=\arctan(x).$$ As far as I know, $$\tan(x)^{-1}$$ can be interpreted as the reciprocal of the tangent, i.e. the cotangent

2017年1月27日 · Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above.

$$\lim_{x\to\infty}\tan^{-1}(\dfrac{x}{4})$$ I have a question about why this limit is $\pi/2$. If the argument of $\tan^{-1}$ goes to infinity, doesn't the slope (since it is the tangent function...

Your series is : $$\arctan(x)=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots$$ (this is easy to obtain from $\ \displaystyle\arctan(x)=\int \frac {dx}{1+x^2}$ using ...

The $\arctan$ function is the inverse function of $$\tan:\left(-\frac{\pi}2,\frac{\pi}2\right)\rightarrow ...