What is arctan (4/3)?
I'm refreshing my memory on sines, cosines, SOHCAHTOA, and writing complex numbers in polar form - I chose the complex number $3+4i$ to try and write it in polar form. The modulus is easily computed to be $5$. However, I'm having some trouble with computing its argument. Based on drawing out the right triangle, thinking of $\tan(\theta)$, then the argument must be $\tan^{-1} (\frac{4}{3})$.
However, how do we actually compute this?
I don't think it's $\frac{1}{\tan(\theta)}$ = $\frac{\cos(\theta)}{\sin(\theta)}$, because that just seems weird and more like cotangent.
Where have I gone wrong?
Thanks,
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$\begingroup$Just a complement to other answers. At first, we have that $\arctan\frac{4}{3}$ is not a rational multiple of $\pi$. Assuming $\arctan\frac{4}{3}\in\pi\mathbb{Q}$ we have that $z=\frac{3+4i}{5}$ is a primitive $q$-th root of unity, but its minimal polynomial over $\mathbb{Q}$ is given by $5z^2-6z+5$, that is not a monic polynomial, hence is not a cyclotomic polynomial. So $\arctan\frac{4}{3}$ is not a "nice" angle. However, there are plenty of ways for approximating values of the arctangent function, for instance by exploiting continued fractions, the (D'Aurizio)-Shafer-Fink inequality, Machin-like formulas or Taylor series. For instance, since
$$ \arctan(z)=\frac{\pi }{4}+\frac{1}{2}(z-1)-\frac{1}{4} (z-1)^2+\frac{1}{12} (z-1)^3-\frac{1}{40} (z-1)^5+\frac{1}{48} (z-1)^6+O((z-1)^7)$$ we have: $$ \arctan\frac{4}{3}\approx \frac{\pi}{4}+\frac{24827}{174960} $$ and this approximation gives us the first five figures $\color{red}{0.92729}$.
$\endgroup$ $\begingroup$The use of the $-1$ in $tan^{-1}\theta$ can be interpreted in two ways.
As an inverse function - this is know as $\arctan\frac43$ and means the angle whose tangent is $\frac43$. For most values (including yours) there is no other way to write it short of writing a decimal approximation obtained from a calculator or similar. $\arctan\frac43\approx0.9272952180\cdots$
As a reciprocal of a function - this is know as cotangent and you gave the correct definition, e.g. $\cot\theta=\frac{\cos\theta}{\sin\theta}$.
Usually the notation $\tan^{-1}\theta$ is reserved for the first meaning.
$\endgroup$ $\begingroup$$\texttt{Newton-Rapson Method}$ yields a good approximation $\displaystyle{~\left(~\mbox{starting from}\ 1 = \tan\left(\,{\pi \over 4}\,\right) ~\right)}$ with $\color{#f00}{\underline{3}}$ iterations: $$ \left\lbrace\begin{array}{rcl} \displaystyle{r_{0}} & \displaystyle{=} & \displaystyle{1} \\[1mm] \displaystyle{r_{n + 1}} & \displaystyle{=} & \displaystyle{r_{n} + {4 \over 3}\,\cos^{2}\left(\, r_{n}\,\right) - {1 \over 2}\,\sin\left(\, 2r_{n}\,\right)\,,\qquad n \geq 0} \end{array}\right. $$
$$ \begin{array}{cl}\hline \displaystyle{n} & \displaystyle{\phantom{111}r_{n}} \\ \hline \displaystyle{0} & \displaystyle{1.000000} \\\displaystyle{1} & \displaystyle{0.934587} \\ \displaystyle{2} & \displaystyle{0.927366} \\ \displaystyle{\color{#f00}{3}} & \displaystyle{0.927295} \\ \displaystyle{4} & \displaystyle{\color{#f00}{0.927295}} \\ \hline \end{array}\qquad\qquad\qquad \left\vert\,{\color{#f00}{0.927295} \over \arctan\left(\, 4/3\,\right)} - 1 \,\right\vert \times 100\ \% \approx 2.35 \times 10^{-5}\ \% $$
$\endgroup$ $\begingroup$Note that $\displaystyle{{4 \over 3} > {\pi \over 4}}$. For this reason, our starting value was $\displaystyle{1 = \tan\left(\,{\pi \over 4}\,\right)}$.
First off, its an angle in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, and $\tan \theta = \dfrac{4}{3}$. It is often times that you are asked to find some thing like $\sin \theta $ for $0 < \theta < \dfrac{\pi}{2}$. With this in mind, you have:
$\theta = \tan^{-1}\left(\dfrac{4}{3}\right)\implies \tan \theta = \dfrac{4}{3}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}\implies \sin \theta = \dfrac{4}{5},\cos \theta = \dfrac{3}{5}$
$\endgroup$ $\begingroup$Oh. NOW I understand your question and confusion.
$\tan^{-1}(4/3)$ in this context means the functional inverse of tangent. If $f$ is a function, such as $\tan$ is, then $f^{-1}$ means the function that maps the "output" of $f$ back to the input of $f$. i.e. If $f(x) = w$ then $f^{-1}(w) = x$.
So $\tan^{-1}(4/3)$ means "the angle, $x$, so that $\tan x = 4/3$". $\tan^{-1}$ is often called the arctangent and is often written $\arctan 4/3$.
This is not to be confused with the multiplicative inverse, which is simply the fractional reciprical; $y^{-1} = \frac 1y$. Unfortunately they are written the exact same @#%!ing way.
In context it should be clear. In figuring the argument for $3 + 4i$ you want the angle so that $\tan x = 4/3$. This is $\arctan 4/3$ or $\tan^{-1} 4/3$. Obviously you do not figure this out by making a fraction out of $\frac 1{\tan 4/3}$. You were right to be confused--- that wouldn't make any freaking sense.
So ... you are probably asking how do you calculate $\arctan 4/3$? Well, there is no standard way. You use tables or a calculator. It's easy to see that it will also be the angle $x$ so that $\sin x = 4/5$ and $\cos x = 3/5$.
Whatever that is.
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How to compute?
$\frac {\sin x}{\cos x} = \frac 43$
So $\sin x = 4k;\cos x = 3k$ for some $k$ and so $(4^2k^2) + (3^2k^2)=1$ so $k = \pm 5$.
So $x = \sin^{-1} \pm 4/5 = \cos^{-1} \pm 3/5 = \tan^{-1} 4/3$.
How do you calculate those angle without a calculator? You don't. It is the base angle of a right triangle with base 3 and height 4. whatever that is.
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