Real Cosine Function is Bounded

Theorem

Let $x \in \R$.

Then:

$\size {\cos x} \le 1$

Proof

From the algebraic definition of the real cosine function:

$\ds \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$

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

Similarly, $\sin x$ is also a real function.

Thus it follows that:

\(\ds \cos^2 x\)

\(\le\)

\(\ds \cos^2 x + \sin^2 x\)

as $\sin^2 x \ge 0$ by Square of Real Number is Non-Negative

\(\ds \)

\(=\)

\(\ds 1\)

Sum of Squares of Sine and Cosine

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

• Complex Cosine Function is Unbounded

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (3) \ \text{(ii)}$