Prove $|\sin(n)|$ or $|\sin(n + 1)|$ > $\sin(\frac{1}{2})$

Question

Let $n$ be a positive integer. Prove that either $|\sin(n)|$ or $|\sin(n + 1)|$ > $\sin(\frac{1}{2})$.

It makes sense to me if I look at the graph of $|\sin x|$. The interval from $k\pi$ to $k\pi+\pi/2$ contains 2 integers. I think if one of them is less than $\sin(\frac{1}{2})$, the other one must be greater than $\sin(\frac{1}{2})$. However, I would like to make it rigorously.

Can we use other technique to prove it? I tried using contradiction (assume both $|\sin(n)|$ and $|\sin(n + 1)|$ < $\sin(\frac{1}{2})$), but I was stuck.

Thanks for your help.

Answer 1

Let's prove this by contradiction: Suppose that neither $|\sin(n)|$ nor $|\sin(n + 1)|$ > $\sin(\frac{1}{2})$, that is to say, suppose that,

$$|\sin{n}| \leq \sin{\frac12} \land |\sin{(n+1)}| \leq \sin{\frac12},$$

where $n\in\mathbb{Z}$.

Squaring the two inequalities and adding them together,

$$\sin^2{n} + \sin^2{(n+1)} \leq 2\sin^2{\frac12}.$$

An equivalent form of the above inequality is:

$$\sin^2{n} + \sin^2{(n+1)} \leq 2\sin^2{\frac12} \iff 2\cos{(1)}\sin^2{(n+\frac12)} \leq 0\\ \implies \sin^2{(n+\frac12)} \leq 0.$$

Since the square of a real number can't be strictly less than zero, the way the inequality $\sin^2{(n+\frac12)} \leq 0$ can be true is if we have equality, i.e., if

$$\sin^2{(n+\frac12)} = 0 \iff \sin{(n+\frac12)} = 0\\ \iff n=k \pi - \frac12,~~~\text{where }k\in\mathbb{Z}.$$

This contradicts are initial assumption $n\in\mathbb{Z}$.

Answer 2

I will solve the base case rest of argument is similiar.

Claim: If $0\leq sin(x) \leq sin(\dfrac{1}{2})$ then $sin(x+1)\geq sin(\dfrac{1}{2})$ where $x\in [0,\pi/2]$

from above constraction we can say that $0\leq x \leq 1/2$

Now, think the unit circle, since $x$ is radian it is equal the corresponding length on circle and $\pi$ is the half of the circle.

case $1$:If $x+1\leq 1/2$

Since $\pi/4\leq1$, $x+1\geq\pi/6$ we are done.

case $2$: Else:

Now, think the corresponding piece with the length $x+1$.

The remaning length $\pi-x-1\geq \pi-3/2 \geq1/2$.

Since the remainig length is bigger, $sin(x+1)\ge sin(1/2)$ as they are $y$ values of this points.

Note: My solution use geometric arguments and and @David 's solution is better.