Sign of Sine

Theorem
For all integer $n$:

where $\sin$ is the real sine function.

Proof
First the case where $n \ge 0$ is addressed.

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:


 * $\forall x \in \R:$


 * $2 n \pi < x < \left({2 n + 1}\right) \pi \implies \sin x > 0$


 * $\left({2 n + 1}\right) \pi < x < \left({2 n + 2}\right) \pi \implies \sin x < 0$

Basis for the Induction
Let $n = 0$.

From the corollary to Sine and Cosine are Periodic on Reals:


 * $\sin x$ is strictly positive on the interval $\left({0 \,.\,.\, \pi}\right)$

and:
 * $\sin x$ is strictly negative on the interval $\left({\pi \,.\,.\, 2 \pi}\right)$

Thus:
 * $2 \cdot 0 \cdot \pi < x < \left({2 \cdot 0 + 1}\right) \pi \implies \sin x > 0$


 * $\left({2 \cdot 0 \cdot + 1}\right) \pi < x < \left({2 \cdot 0 + 2}\right) \pi \implies \sin x < 0$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:


 * $\forall x \in \R:$


 * $2 k \pi < x < \left({2 k + 1}\right) \pi \implies \sin x > 0$


 * $\left({2 k + 1}\right) \pi < x < \left({2 k + 2}\right) \pi \implies \sin x < 0$

Then we need to show:


 * $\forall x \in \R:$


 * $2 \left({k + 1}\right) \pi < x < \left({2 \left({k + 1}\right) + 1}\right) \pi \implies \sin x > 0$


 * $\left({2 \left({k + 1}\right) + 1}\right) \pi < x < \left({2 \left({k + 1}\right) + 2}\right) \pi \implies \sin x < 0$

That is:


 * $\forall x \in \R:$


 * $\left({2 k + 2}\right) \pi < x < \left({2 k + 3}\right) \pi \implies \sin x > 0$


 * $\left({2 k + 3}\right) \pi < x < \left({2 k + 4}\right) \pi \implies \sin x < 0$

Induction Step
This is our induction step:

Also:

It follows by induction that:


 * $\forall n \in \Z_{\ge 0}: \forall x \in \R:$


 * $2 n \pi < x < \left({2 n + 1}\right) \pi \implies \sin x > 0$


 * $\left({2 n + 1}\right) \pi < x < \left({2 n + 2}\right) \pi \implies \sin x < 0$

Negative $n$
Let $n \in \Z_{\le 0}$ be a negative integer.

Then, by definition, $- \left({n + 1}\right)$ is a (strictly) positive integer.

So:

Similarly:

Also see

 * Sign of Cosine
 * Sign of Tangent
 * Sign of Cosecant
 * Sign of Secant
 * Sign of Cotangent