Solutions of sin x equals sin a

Theorem

Let $\alpha \in \closedint {-1} 1$ be fixed.

Let:

$(1): \quad \sin x = \sin \alpha$

The solution set of $(1)$ is:

$\set {x \in \R: \forall n \in \Z: x = n \pi + \paren {-1}^n \alpha}$

$\map \sin {\pi - x} = \sin x$

$\ds x$

$=$

$\ds 2 n \pi + a$

$\ds x$

$=$

$\ds \paren {2 n + 1} \pi - a$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds n \pi + \paren {-1}^n x$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(34)$