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:


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

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 Order of Squares in Totally Ordered Ring 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.