Sign of Sine
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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 $n = 0$.
From the corollary to Sine and Cosine are Periodic on Reals:
- $\sin x$ 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 |
$\blacksquare$