Real Cosine Function is Bounded

Theorem
Let $x \in \R$.

Then:
 * $\left|{\cos x}\right| \le 1$
 * $\left|{\sin x}\right| \le 1$

Proof
When $x \in \R$ it follows from the algebraic definitions for sine and cosine:


 * $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n+1}}{\left({2n+1}\right)!}$
 * $\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2n}}{\left({2n}\right)!}$

that $\sin x$ and $\cos x$ are real functions.

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

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

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

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

The result follows.

Note
This result holds only for real values of $x$.

When $x$ is complex, this result does not apply.