Sign of Sine

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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 $n = 0$.

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:

\(\displaystyle 2 k \pi < x < \left({2 k + 1}\right) \pi\) \(\implies\) \(\displaystyle \sin x > 0\)
\(\displaystyle \therefore \ \ \) \(\displaystyle 2 k \pi < \left({x - 2 \pi}\right) < \left({2 k + 1}\right) \pi\) \(\implies\) \(\displaystyle \sin \left({x - 2 \pi}\right) > 0\) 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

$\blacksquare$


Also see