I was trying to prove following inequality:

$$|\sin n\theta| \leq n\sin \theta \ \text{for all n=1,2,3... and } \ 0<\theta<π $$

I succeeded in proving this via induction but I didn't get "feel" over the proof. Are there other proof for this inequality?


3 Answers 3


Not sure that this is what you want, but a neat way to do it is noticing that if $0 < \theta < \pi$:

$|1+e^{2i\theta}+...+e^{2i(n-1)\theta}|=\frac{|\sin (n\theta)|}{\sin (\theta)}$ and then use the triangle inequality on LHS


You can show that $|\sin(x)|$ is subadditive, i.e. $$|\sin(x + y)| \le |\sin(x)| + |\sin(y)|.$$ To prove this, simply expand the left side: \begin{align*} |\sin(x + y)| &= |\sin(x)\cos(y) + \sin(y)\cos(x)| \\ &\le |\sin(x)| \cdot |\cos(y)| + |\sin(y)| \cdot | \cos(y)| \\ &\le |\sin(x)| + |\sin(y)|, \end{align*} as $|\cos(x)|$ and $|\cos(y)|$ are less than or equal to $1$.

How does this help? Note that, when $0 < \theta < \pi$, we have $\sin(x) \ge 0$, hence $|\sin(\theta)| = \sin(\theta)$. Using induction, we can use subadditivitiy to show that $$|\sin(n\theta)| = |\sin(\underbrace{\theta + \theta + \ldots + \theta}_{\text{n times}})| \le \underbrace{|\sin(\theta)| + \ldots + |\sin(\theta)|}_{\text{n times}} = n|\sin(\theta)| = n \sin(\theta).$$

  • 2
    $\begingroup$ Note that OP already “succeeded in proving this via induction” and asks for an alternative proof. $\endgroup$
    – Martin R
    Sep 13, 2019 at 7:29

Let $z = \cos(θ) + i \sin(θ) $

$\displaystyle {\sin(n\,θ)\over \sin(θ)} = {\Im(z^n) \over \Im(z)} = {z^n - 1/z^n \over z - 1/z} = z^{n-1} + z^{n-3}+\cdots+{1\over z^{n-3}} + {1\over z^{n-1}}$

RHS can be group in pairs (except possibly center 1, if n is odd) :

${\displaystyle z^k + {1 \over z^k} = 2 \cos(k\,x) }$

Sum of pair absolute value at most reached 2.

With RHS having n terms, $\displaystyle \left|{\sin(n\,θ)\over \sin(θ)} \right| \le n$


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