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: {{begin-eqn}} {{eqn | l = 2 k \pi < x < \paren {2 k + 1} \pi       | o = \implies       | r = \sin x > 0 }} {{eqn | ll= \leadsto       | l = 2 k \pi < \paren {x - 2 \pi} < \paren {2 k + 1} \pi       | o = \implies       | r = \map \sin {x - 2 \pi} > 0       | c = replacing $x$ with $x - 2 \pi$ }} {{eqn | ll= \leadsto       | l = 2 k \pi < \paren {x - 2 \pi} < \paren {2 k + 1} \pi       | o = \implies       | r = \sin x > 0       | c = Sine and Cosine are Periodic on Reals }} {{eqn | ll= \leadsto       | l = \paren {2 k + 2} \pi < x < \paren {2 k + 3} \pi       | o = \implies       | r = \sin x > 0       | c = adding $2 \pi$ to all elements of inequality }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \paren {2 k + 1} \pi < x < \paren {2 k + 2} \pi       | o = \implies       | r = \sin x < 0 }} {{eqn | ll= \leadsto       | l = \paren {2 k + 1} \pi < \paren {x - 2 \pi} < \paren {2 k + 2} \pi       | o = \implies       | r = \map \sin {x - 2 \pi} < 0       | c = replacing $x$ with $x - 2 \pi$ }} {{eqn | ll= \leadsto       | l = \paren {2 k + 1} \pi < \paren {x - 2 \pi} < \paren {2 k + 2} \pi       | o = \implies       | r = \sin x < 0       | c = Sine and Cosine are Periodic on Reals }} {{eqn | ll= \leadsto       | l = \paren {2 k + 3} \pi < x < \paren {2 k + 4} \pi       | o = \implies       | r = \sin x < 0       | c = adding $2 \pi$ to all elements of inequality }} {{end-eqn}} It follows by induction that: :$\forall n \in \Z_{\ge 0}: \forall x \in \R:$ ::$2 n \pi < x < \paren {2 n + 1} \pi \implies \sin x > 0$ ::$\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi \implies \sin x < 0$ {{qed|lemma}} === Negative $n$ === Let $n \in \Z_{\le 0}$ be a negative integer. Then, by definition, $-\paren {n + 1}$ is a (strictly) positive integer. So: {{begin-eqn}} {{eqn | l = 2 \paren {-\paren {n + 1} } \pi < x < \paren {2 \paren {-\paren {n + 1} } + 1} \pi       | o = \implies       | r = \sin x > 0       | c = Result for positive $n$ above }} {{eqn | l = -\paren {2 n + 2} \pi < x < -\paren {2 n + 1} \pi       | o = \implies       | r = \sin x > 0       | c = simplifying }} {{eqn | ll= \leadsto       | l = -\paren {2 n + 2} \pi < \paren {-x} < -\paren {2 n + 1} \pi       | o = \implies       | r = \map \sin {-x} > 0       | c = replacing $x$ with $-x$ }} {{eqn | ll= \leadsto       | l = -\paren {2 n + 2} \pi < \paren {-x} < -\paren {2 n + 1} \pi       | o = \implies       | r = -\paren {\sin x} > 0       | c = Sine Function is Odd }} {{eqn | ll= \leadsto       | l = \paren {2 n + 2} \pi > x > \paren {2 n + 1} \pi       | o = \implies       | r = \sin x < 0       | c = multiplying throughout by $-1$ }} {{eqn | ll= \leadsto       | l = \paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi       | o = \implies       | r = \sin x < 0       | c = rearranging }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = \paren {2 \paren {-\paren {n + 1} } + 1} \pi < x < \paren {2 \paren {-\paren {n + 1} } + 2} \pi       | o = \implies       | r = \sin x < 0       | c = Result for positive $n$ above }} {{eqn | l = -\paren {2 n + 1} \pi < x < -2 n \pi       | o = \implies       | r = \sin x < 0       | c = simplifying }} {{eqn | ll= \leadsto       | l = -\paren {2 n + 1} \pi < \paren {-x} < -2 n \pi       | o = \implies       | r = \map \sin {-x} < 0       | c = replacing $x$ with $-x$ }} {{eqn | ll= \leadsto       | l = -\paren {2 n + 1} \pi < \paren {-x} < -2 n \pi       | o = \implies       | r = -\paren {\sin x} < 0       | c = Sine Function is Odd }} {{eqn | ll= \leadsto       | l = -\paren {2 n + 1} \pi > x > 2 n \pi       | o = \implies       | r = \sin x > 0       | c = multiplying throughout by $-1$ }} {{eqn | ll= \leadsto       | l = 2 n \pi < x < \paren {2 n + 1} \pi       | o = \implies       | r = \sin x > 0       | c = simplifying and rearranging }} {{end-eqn}} {{qed}}