Solutions of sin x equals sin a/Proof
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Theorem
Let $\alpha \in \closedint {-1} 1$ be fixed.
Let:
- $(1): \quad \sin x = \sin \alpha$
The solution set of $(1)$ is:
- $\set {x \in \R: \forall n \in \Z: x = n \pi + \paren {-1}^n \alpha}$
Proof
From Sine of Supplementary Angle:
- $\map \sin {\pi - x} = \sin x$
and so from Real Sine Function is Periodic:
\(\ds x\) | \(=\) | \(\ds 2 n \pi + a\) | ||||||||||||
\(\ds x\) | \(=\) | \(\ds \paren {2 n + 1} \pi - a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds n \pi + \paren {-1}^n x\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: General solution of $\sin x = \sin \alpha$