Question

Cos^-1 (-x) = ? please explain the answer?

Answer 1

Answer on Question# #50656 – Mathematics – Trigonometry

Question:

Cos^ -1 (-x) = ? Please explain the answer.

Solution:

Let us write some definitions.

A function $f$ is said to be an even function if for any number $x$, $f(-x) = f(x)$.

A function $f$ is said to be an odd function if for any number $x$, $f(-x) = -f(x)$.

A function $\cos^{-1}(x)$ (or $\arccos(x)$, it is usually called arccosine function) is the inverse cosine function, defined to be the inverse of the restricted cosine function $\cos(x)$ at interval $0 \leq x \leq \pi$.

Arccosine is neither even nor odd function:

$$\arccos s(-x) \neq \pm \arccos s(x).$$

Let us show it:

$$\arccos s(-x) = \arccos s(-\cos(\arccos(x))) = \arccos s(\cos(\pi - \arccos(x))) = \pi - \arccos(x)$$

Here we used the following relations:

$$\cos(\pi - x) = -\cos(x),$$ $$\cos(\arccos(x)) = x, \quad \text{when } -1 \leq x \leq 1,$$ $$\arccos(\cos(y)) = y, \quad \text{when } 0 \leq y \leq \pi.$$

Using notation $\cos^{-1}(x)$, the left-hand and right-hand sides of (1) give the following equality:

$$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$$

Answer: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.

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