Real Sine Function is Bounded

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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.

$\blacksquare$


Also see


Sources