Sine of Half-Integer Multiple of Pi

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Theorem

Let $x \in \R$ be a real number.

Let $\sin x$ be the sine of $x$.


Then:

$\forall n \in \Z: \map \sin {n + \dfrac 1 2} \pi = \paren {-1}^n$

or:

\(\displaystyle \forall m \in \Z: \ \ \) \(\displaystyle \map \sin {2 m + \dfrac 1 2} \pi\) \(=\) \(\displaystyle 1\)
\(\displaystyle \forall m \in \Z: \ \ \) \(\displaystyle \map \sin {2 m - \dfrac 1 2} \pi\) \(=\) \(\displaystyle -1\)


Proof

From the discussion of Sine and Cosine are Periodic on Reals, we have that:

$\map \sin {x + \dfrac \pi 2} = \cos x$

The result then follows directly from the Cosine of Multiple of Pi.

$\blacksquare$