Proof by induction: === Base case === For $n = 0$, it follows from Sine and Cosine are Periodic on Reals/Corollary. === Induction Hypothesis === This is our induction hypothesis: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{end-eqn}} Now we need to show true for $n=k+1$: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 3 2} \pi < x < \paren {2 n + \dfrac 5 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 5 2} \pi < x < \paren {2 n + \dfrac 7 2} \pi$ }} {{end-eqn}} === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 3 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 5 2} \pi$ }} {{eqn | l = \map \cos {x + 2 \pi}       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 3 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 5 2} \pi$       | cc = Sine and Cosine are Periodic on Reals }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 3 2} \pi < x < \paren {2 n + \dfrac 5 2} \pi$       | cc = Replacing $x + 2 \pi$ by $x$ }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 5 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 7 2} \pi$ }} {{eqn | l = \map \cos {x + 2 \pi}       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 5 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 7 2} \pi$       | cc = Sine and Cosine are Periodic on Reals }} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 5 2} \pi < x < \paren {2 n + \dfrac 7 2} \pi$       | cc = Replacing $x + 2 \pi$ by $x$ }} {{end-eqn}} The result follows by induction. For negative $n$: === Induction Hypothesis === This is our induction hypothesis: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{end-eqn}} Now we need to show true for $n=k-1$: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 5 2} \pi < x < \paren {2 n - \dfrac 3 2} \pi$ }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 3 2} \pi < x < \paren {2 n - \dfrac 1 2} \pi$ }} {{end-eqn}} === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \map \cos {x + 2 \pi}       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 1 2} \pi$       | cc= replacing $x$ by $x + 2 \pi$ }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 1 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 1 2} \pi$       | cc= Sine and Cosine are Periodic on Reals }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 5 2} \pi < x < \paren {2 n - \dfrac 3 2} \pi$ }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{eqn | l = \map \cos {x + 2 \pi}       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 3 2} \pi$       | cc= replacing $x$ by $x + 2 \pi$ }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n + \dfrac 1 2} \pi < x + 2 \pi < \paren {2 n + \dfrac 3 2} \pi$       | cc= Sine and Cosine are Periodic on Reals }} {{eqn | l = \cos x       | o = >       | r = 0       | c = for $\paren {2 n - \dfrac 3 2} \pi < x < \paren {2 n - \dfrac 1 2} \pi$ }} {{end-eqn}} The result follows by induction. {{qed}}