# Sign of Sine

## Theorem

For all integer $n$:

 $\displaystyle \sin x$ $>$ $\displaystyle 0$ for all real $x$ such that $2 n \pi < x < \left({2 n + 1}\right) \pi$ $\displaystyle \sin x$ $<$ $\displaystyle 0$ for all real $x$ such that $\left({2 n + 1}\right) \pi < x < \left({2 n + 2}\right) \pi$

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 $n = 0$.

$\sin x$ 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:

 $\displaystyle 2 k \pi < x < \left({2 k + 1}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ $\displaystyle \therefore \ \$ $\displaystyle 2 k \pi < \left({x - 2 \pi}\right) < \left({2 k + 1}\right) \pi$ $\implies$ $\displaystyle \sin \left({x - 2 \pi}\right) > 0$ replacing $x$ with $x - 2 \pi$ $\displaystyle \therefore \ \$ $\displaystyle 2 k \pi < \left({x - 2 \pi}\right) < \left({2 k + 1}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ Sine and Cosine are Periodic on Reals $\displaystyle \therefore \ \$ $\displaystyle \left({2 k + 2}\right) \pi < x < \left({2 k + 3}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ adding $2 \pi$ to all elements of inequality

Also:

 $\displaystyle \left({2 k + 1}\right) \pi < x < \left({2 k + 2}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ $\displaystyle \therefore \ \$ $\displaystyle \left({2 k + 1}\right) \pi < \left({x - 2 \pi}\right) < \left({2 k + 2}\right) \pi$ $\implies$ $\displaystyle \sin \left({x - 2 \pi}\right) < 0$ replacing $x$ with $x - 2 \pi$ $\displaystyle \therefore \ \$ $\displaystyle \left({2 k + 1}\right) \pi < \left({x - 2 \pi}\right) < \left({2 k + 2}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ Sine and Cosine are Periodic on Reals $\displaystyle \therefore \ \$ $\displaystyle \left({2 k + 3}\right) \pi < x < \left({2 k + 4}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ adding $2 \pi$ to all elements of inequality

It follows by induction that:

$\forall n \in \Z_{\ge 0}: \forall x \in \R:$
$2 n \pi < x < \left({2 n + 1}\right) \pi \implies \sin x > 0$
$\left({2 n + 1}\right) \pi < x < \left({2 n + 2}\right) \pi \implies \sin x < 0$

$\Box$

### 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:

 $\displaystyle 2 \left({- \left({n + 1}\right)}\right) \pi < x < \left({2 \left({- \left({n + 1}\right)}\right) + 1}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ Result for positive $n$ above $\displaystyle -\left({2 n + 2}\right) \pi < x < -\left({2 n + 1}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ simplifying $\displaystyle \therefore \ \$ $\displaystyle -\left({2 n + 2}\right) \pi < \left({-x}\right) < -\left({2 n + 1}\right) \pi$ $\implies$ $\displaystyle \sin \left({-x}\right) > 0$ replacing $x$ with $-x$ $\displaystyle \therefore \ \$ $\displaystyle -\left({2 n + 2}\right) \pi < \left({-x}\right) < -\left({2 n + 1}\right) \pi$ $\implies$ $\displaystyle -\left({\sin x}\right) > 0$ Sine Function is Odd $\displaystyle \therefore \ \$ $\displaystyle \left({2 n + 2}\right) \pi > x > \left({2 n + 1}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ multiplying throughout by $-1$ $\displaystyle \therefore \ \$ $\displaystyle \left({2 n + 1}\right) \pi < x < \left({2 n + 2}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ rearranging

Similarly:

 $\displaystyle \left({2 \left({- \left({n + 1}\right)}\right) + 1}\right) \pi < x < \left({2 \left({- \left({n + 1}\right)}\right) + 2}\right) \pi$ $\implies$ $\displaystyle \sin x < 0$ Result for positive $n$ above $\displaystyle -\left({2 n + 1}\right) \pi < x < -2 n \pi$ $\implies$ $\displaystyle \sin x < 0$ simplifying $\displaystyle \therefore \ \$ $\displaystyle -\left({2 n + 1}\right) \pi < \left({-x}\right) < -2 n \pi$ $\implies$ $\displaystyle \sin \left({-x}\right) < 0$ replacing $x$ with $-x$ $\displaystyle \therefore \ \$ $\displaystyle -\left({2 n + 1}\right) \pi < \left({-x}\right) < -2 n \pi$ $\implies$ $\displaystyle -\left({\sin x}\right) < 0$ Sine Function is Odd $\displaystyle \therefore \ \$ $\displaystyle -\left({2 n + 1}\right) \pi > x > 2 n \pi$ $\implies$ $\displaystyle \sin x > 0$ multiplying throughout by $-1$ $\displaystyle \therefore \ \$ $\displaystyle 2 n \pi < x < \left({2 n + 1}\right) \pi$ $\implies$ $\displaystyle \sin x > 0$ simplifying and rearranging

$\blacksquare$