Why is $\frac{\sin (-x)}{\cos (-x)}$ not equal to $\frac{-\sin (x)}{-\cos (x)}$

Question

This is the first scenario I have seen of math semi-contradicting its rules. By the negative trig identity definition, shouldn't $-\sin(x) = \sin(-x)$? Therefore, why is the title statement wrong?

Answer 1

The core reason is tied to the notion of an even/odd function.

An even function $f$ satisfies the condition $f(-x) = f(x)$ for all $x$.

An odd function $f$ satisfies the condition $f(-x) = -f(x)$ for all $x$.

Sine is an odd function and cosine is even. Therefore:

$$\frac{\sin(-x)}{\cos(-x)} = \frac{-\sin(x)}{\cos(x)}$$

Notice this differs by a factor of $(-1)$ from your proposed solution.

Answer 2

The function $ f(x) = \sin(x) $ is an odd function because $f$'s graph is symmetric with respect to the origin $(0,0)$. $f$ is odd if $f(-x) = -f(x)$.

In the graph below, the blue curve depicts $f(x) = \sin(x)$, which clearly is symmetric with respect to the origin. Thus, $f(-x) = \sin(-x) = -\sin(x)$.

On the other hand, the function $ g(x) = \cos(x) $ is an even function because $g$'s graph is symmetric with respect to the $y$-axis. $g$ is even if $g(-x) = g(x)$.

In the graph below, the red curve depicts $g(x) = \cos(x)$, which clearly is symmetric with respect to the $y$-axis. Thus, $g(-x) = \cos(-x) = \cos(x)$.

Hence $$\frac{\sin(-x)}{\cos(-x)} \neq \frac{-\sin(x)}{-\cos(x)}$$