# angle between arcs

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##### 1: 1.9 Calculus of a Complex Variable
and when $z\neq 0$, …
###### Conformal Transformation
The angle between $C_{1}$ and $C_{2}$ at $z_{0}$ is the angle between the tangents to the two arcs at $z_{0}$, that is, the difference of the signed angles that the tangents make with the positive direction of the real axis. If $f^{\prime}(z_{0})\not=0$, then the angle between $C_{1}$ and $C_{2}$ equals the angle between $C^{\prime}_{1}$ and $C^{\prime}_{2}$ both in magnitude and sense. We then say that the mapping $w=f(z)$ is conformal (angle-preserving) at $z_{0}$. …
##### 2: 4.42 Solution of Triangles
4.42.10 $\sin a\cos B=\cos b\sin c-\sin b\cos c\cos A,$
##### 3: 31.16 Mathematical Applications
Expansions of Heun polynomial products in terms of Jacobi polynomial (§18.3) products are derived in Kalnins and Miller (1991a, b, 1993) from the viewpoint of interrelation between two bases in a Hilbert space:
31.16.1 $\mathit{Hp}_{n,m}\left(x\right)\mathit{Hp}_{n,m}\left(y\right)=\sum_{j=0}^{n}A% _{j}{\sin}^{2j}\theta\*P^{(\gamma+\delta+2j-1,\epsilon-1)}_{n-j}\left(\cos% \left(2\theta\right)\right)P^{(\delta-1,\gamma-1)}_{j}\left(\cos\left(2\phi% \right)\right),$
##### 4: 7.20 Mathematical Applications
Then the arc length between the origin and $P(t)$ equals $t$, and is directly proportional to the curvature at $P(t)$, which equals $\pi t$. Furthermore, because $\ifrac{\mathrm{d}y}{\mathrm{d}x}=\tan\left(\frac{1}{2}\pi t^{2}\right)$, the angle between the $x$-axis and the tangent to the spiral at $P(t)$ is given by $\frac{1}{2}\pi t^{2}$. …
##### 5: 1.6 Vectors and Vector-Valued Functions
###### Magnitude and Angle of Vector $\mathbf{a}$
1.6.4 $\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{\left\|{\mathbf{a}}\right\|\;\left% \|{\mathbf{b}}\right\|};$
$\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. …
1.6.9 $\mathbf{a}\times\mathbf{b}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ a_{1}&a_{2}&a_{3}\\ b_{1}&b_{2}&b_{3}\end{vmatrix}\\ =(a_{2}b_{3}-a_{3}b_{2})\mathbf{i}+(a_{3}b_{1}-a_{1}b_{3})\mathbf{j}+(a_{1}b_{% 2}-a_{2}b_{1})\mathbf{k}\\ =\left\|{\mathbf{a}}\right\|\left\|{\mathbf{b}}\right\|(\sin\theta)\mathbf{n},$
For a sphere $x=\rho\sin\theta\cos\phi$, $y=\rho\sin\theta\sin\phi$, $z=\rho\cos\theta$, …
19.11.2 $E\left(\theta,k\right)+E\left(\phi,k\right)=E\left(\psi,k\right)+k^{2}\sin% \theta\sin\phi\sin\psi.$
$\sin\psi=\frac{(\sin\theta\cos\phi)\Delta(\phi)+(\sin\phi\cos\theta)\Delta(% \theta)}{1-k^{2}{\sin}^{2}\theta{\sin}^{2}\phi},$
19.11.14 $\sin\psi=(\sin 2\theta)\Delta(\theta)/(1-k^{2}{\sin}^{4}\theta),$
##### 7: 20.15 Tables
This reference gives $\theta_{j}\left(x,q\right)$, $j=1,2,3,4$, and their logarithmic $x$-derivatives to 4D for $x/\pi=0(.1)1$, $\alpha=0(9^{\circ})90^{\circ}$, where $\alpha$ is the modular angle given by
20.15.1 $\sin\alpha={\theta_{2}}^{2}\left(0,q\right)/{\theta_{3}}^{2}\left(0,q\right)=k.$
31.10.8 ${\sin}^{2}\theta\left(\frac{{\partial}^{2}\mathcal{K}}{{\partial\theta}^{2}}+% \left((1-2\gamma)\tan\theta+2(\delta+\epsilon-\tfrac{1}{2})\cot\theta\right)% \frac{\partial\mathcal{K}}{\partial\theta}-4\alpha\beta\mathcal{K}\right)+% \frac{{\partial}^{2}\mathcal{K}}{{\partial\phi}^{2}}+\left((1-2\delta)\cot\phi% -(1-2\epsilon)\tan\phi\right)\frac{\partial\mathcal{K}}{\partial\phi}=0.$
31.10.9 $\mathcal{K}(\theta,\phi)=P\begin{Bmatrix}0&1&\infty&\\[1.0pt] 0&\frac{1}{2}-\delta-\sigma&\alpha&{\cos}^{2}\theta\\[1.0pt] 1-\gamma&\frac{1}{2}-\epsilon+\sigma&\beta&\end{Bmatrix}\*P\begin{Bmatrix}0&1&% \infty&\\[1.0pt] 0&0&-\frac{1}{2}+\delta+\sigma&{\cos}^{2}\phi\\[1.0pt] 1-\epsilon&1-\delta&-\frac{1}{2}+\epsilon-\sigma&\end{Bmatrix},$
31.10.21 $\frac{{\partial}^{2}\mathcal{K}}{{\partial r}^{2}}+\frac{2(\gamma+\delta+% \epsilon)-1}{r}\frac{\partial\mathcal{K}}{\partial r}+\frac{1}{r^{2}}\frac{{% \partial}^{2}\mathcal{K}}{{\partial\theta}^{2}}+\frac{(2(\delta+\epsilon)-1)% \cot\theta-(2\gamma-1)\tan\theta}{r^{2}}\frac{\partial\mathcal{K}}{\partial% \theta}+\frac{1}{r^{2}{\sin}^{2}\theta}\frac{{\partial}^{2}\mathcal{K}}{{% \partial\phi}^{2}}+\frac{(2\delta-1)\cot\phi-(2\epsilon-1)\tan\phi}{r^{2}{\sin% }^{2}\theta}\frac{\partial\mathcal{K}}{\partial\phi}=0.$
31.10.22 $\mathcal{K}(r,\theta,\phi)=r^{m}{\sin}^{2p}\theta P\begin{Bmatrix}0&1&\infty&% \\ 0&0&a&{\cos}^{2}\theta\\ \tfrac{1}{2}(3-\gamma)&c&b&\end{Bmatrix}\*P\begin{Bmatrix}0&1&\infty&\\ 0&0&a^{\prime}&{\cos}^{2}\phi\\ 1-\epsilon&1-\delta&b^{\prime}&\end{Bmatrix},$