Trig Functions — Question 10

Question 10 - Solution

(a) Simplify using identities:

Using the identity $$1 - \cos(2x) = 2\sin^2(x),$$ we obtain $$f(x) = \frac{2\sin^2(x)}{\sin(x)}.$$

For $\sin(x) \neq 0$, this simplifies to $$f(x) = 2\sin(x).$$

$$\boxed{f(x) = 2\sin(x) \quad \text{for } \sin(x) \neq 0}$$

(b) Solve $f(x) = 2\sin(x)$

From part (a), the equation $$f(x) = 2\sin(x)$$ is an identity for all $x$ where $f(x)$ is defined.

However, the original function is undefined when $$\sin(x) = 0 \Rightarrow x = 0,\ \pi.$$

Since the interval is restricted to $(0, \pi)$, neither excluded value lies in the interval. Therefore, all values in $(0, \pi)$ satisfy the equation.

$$\boxed{\text{Solution set: } (0, \pi) \quad \text{(excluded: none within the interval)}}$$

(c) Domain of $f(x) = \frac{1 - \cos(2x)}{\sin(x)}$ on $[0, 2\pi]$

The function is undefined when the denominator is zero: $$\sin(x) = 0 \Rightarrow x = 0,\ \pi,\ 2\pi.$$

Removing these values from the interval $[0, 2\pi]$, the domain is $$\boxed{(0, \pi) \cup (\pi, 2\pi)}.$$

$$\text{Excluded values: } x = 0,\ \pi,\ 2\pi$$