Boundedness of Sine and Cosine

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:

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

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.

$\blacksquare$

Note

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

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

Sources

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