Real Sine Function is Bounded

Theorem
Let $x \in \R$.

Then:
 * $\size {\sin x} \le 1$

Proof
From the algebraic definition of the real sine function:


 * $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}$

it follows that $\sin x$ is a real function.

Thus $\sin^2 x \ge 0$.

From Sum of Squares of Sine and Cosine‎, we have that $\cos^2 x + \sin^2 x = 1$.

Thus it follows that:
 * $\sin^2 x = 1 - \cos^2 x \le 1$

From Ordering of Squares in Reals and the definition of absolute value, we have that:
 * $x^2 \le 1 \iff \size x \le 1$

The result follows.

Also see

 * Complex Sine Function is Unbounded