# Sign of Sine

## Theorem

Let $x$ be a real number.

 $\ds \sin x$ $>$ $\ds 0$ if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$ $\ds \sin x$ $<$ $\ds 0$ if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$

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 $\map P n$ be the proposition:

$\forall x \in \R:$
$2 n \pi < x < \paren {2 n + 1} \pi \implies \sin x > 0$
$\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi \implies \sin x < 0$

### Basis for the Induction

Let $n = 0$.

$\sin x$ is strictly positive on the interval $\openint 0 \pi$

and:

$\sin x$ is strictly negative on the interval $\openint \pi {2 \pi}$

Thus:

$2 \cdot 0 \cdot \pi < x < \paren {2 \cdot 0 + 1} \pi \implies \sin x > 0$
$\paren {2 \cdot 0 \cdot + 1} \pi < x < \paren {2 \cdot 0 + 2} \pi \implies \sin x < 0$

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\forall x \in \R:$
$2 k \pi < x < \paren {2 k + 1} \pi \implies \sin x > 0$
$\paren {2 k + 1} \pi < x <\paren {2 k + 2} \pi \implies \sin x < 0$

Then we need to show:

$\forall x \in \R:$
$2 \paren {k + 1} \pi < x < \paren {2 \paren {k + 1} + 1} \pi \implies \sin x > 0$
$\paren {2 \paren {k + 1} + 1} \pi < x < \paren {2 \paren {k + 1} + 2} \pi \implies \sin x < 0$

That is:

$\forall x \in \R:$
$\paren {2 k + 2} \pi < x < \paren {2 k + 3} \pi \implies \sin x > 0$
$\paren {2 k + 3} \pi < x < \paren {2 k + 4} \pi \implies \sin x < 0$

### Induction Step

This is our induction step:

 $\ds 2 k \pi < x < \paren {2 k + 1} \pi$ $\implies$ $\ds \sin x > 0$ $\ds \leadsto \ \$ $\ds 2 k \pi < \paren {x - 2 \pi} < \paren {2 k + 1} \pi$ $\implies$ $\ds \map \sin {x - 2 \pi} > 0$ replacing $x$ with $x - 2 \pi$ $\ds \leadsto \ \$ $\ds 2 k \pi < \paren {x - 2 \pi} < \paren {2 k + 1} \pi$ $\implies$ $\ds \sin x > 0$ Sine and Cosine are Periodic on Reals $\ds \leadsto \ \$ $\ds \paren {2 k + 2} \pi < x < \paren {2 k + 3} \pi$ $\implies$ $\ds \sin x > 0$ adding $2 \pi$ to all elements of inequality

Also:

 $\ds \paren {2 k + 1} \pi < x < \paren {2 k + 2} \pi$ $\implies$ $\ds \sin x < 0$ $\ds \leadsto \ \$ $\ds \paren {2 k + 1} \pi < \paren {x - 2 \pi} < \paren {2 k + 2} \pi$ $\implies$ $\ds \map \sin {x - 2 \pi} < 0$ replacing $x$ with $x - 2 \pi$ $\ds \leadsto \ \$ $\ds \paren {2 k + 1} \pi < \paren {x - 2 \pi} < \paren {2 k + 2} \pi$ $\implies$ $\ds \sin x < 0$ Sine and Cosine are Periodic on Reals $\ds \leadsto \ \$ $\ds \paren {2 k + 3} \pi < x < \paren {2 k + 4} \pi$ $\implies$ $\ds \sin x < 0$ adding $2 \pi$ to all elements of inequality

It follows by induction that:

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

$\Box$

### Negative $n$

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

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

So:

 $\ds 2 \paren {-\paren {n + 1} } \pi < x < \paren {2 \paren {-\paren {n + 1} } + 1} \pi$ $\implies$ $\ds \sin x > 0$ Result for positive $n$ above $\ds -\paren {2 n + 2} \pi < x < -\paren {2 n + 1} \pi$ $\implies$ $\ds \sin x > 0$ simplifying $\ds \leadsto \ \$ $\ds -\paren {2 n + 2} \pi < \paren {-x} < -\paren {2 n + 1} \pi$ $\implies$ $\ds \map \sin {-x} > 0$ replacing $x$ with $-x$ $\ds \leadsto \ \$ $\ds -\paren {2 n + 2} \pi < \paren {-x} < -\paren {2 n + 1} \pi$ $\implies$ $\ds -\paren {\sin x} > 0$ Sine Function is Odd $\ds \leadsto \ \$ $\ds \paren {2 n + 2} \pi > x > \paren {2 n + 1} \pi$ $\implies$ $\ds \sin x < 0$ multiplying throughout by $-1$ $\ds \leadsto \ \$ $\ds \paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ $\implies$ $\ds \sin x < 0$ rearranging

Similarly:

 $\ds \paren {2 \paren {-\paren {n + 1} } + 1} \pi < x < \paren {2 \paren {-\paren {n + 1} } + 2} \pi$ $\implies$ $\ds \sin x < 0$ Result for positive $n$ above $\ds -\paren {2 n + 1} \pi < x < -2 n \pi$ $\implies$ $\ds \sin x < 0$ simplifying $\ds \leadsto \ \$ $\ds -\paren {2 n + 1} \pi < \paren {-x} < -2 n \pi$ $\implies$ $\ds \map \sin {-x} < 0$ replacing $x$ with $-x$ $\ds \leadsto \ \$ $\ds -\paren {2 n + 1} \pi < \paren {-x} < -2 n \pi$ $\implies$ $\ds -\paren {\sin x} < 0$ Sine Function is Odd $\ds \leadsto \ \$ $\ds -\paren {2 n + 1} \pi > x > 2 n \pi$ $\implies$ $\ds \sin x > 0$ multiplying throughout by $-1$ $\ds \leadsto \ \$ $\ds 2 n \pi < x < \paren {2 n + 1} \pi$ $\implies$ $\ds \sin x > 0$ simplifying and rearranging

$\blacksquare$