Trig Functions — Question 5

Question 5 - Solution

Simplify with the domain retained. Half-angle identities give

$$\boxed{f(x)=\tan(x/2)},\qquad x\ne(2k+1)\pi,\ k\in\mathbb Z.$$

Indeed $\sin x=2\sin(x/2)\cos(x/2)$ and $1+\cos x=2\cos^2(x/2)$.

The alternative $\csc x-\cot x$ is valid only when $\sin x\ne0$ and loses valid inputs such as $x=0$.

Solve. $\tan(x/2)=1$ implies $x=\pi/2+2k\pi$, so on $(0,2\pi)$,

$$\boxed{x=\pi/2}.$$

Domain and parity. The original denominator vanishes precisely at odd multiples of $\pi$, hence

$$\boxed{D=\mathbb R\setminus\{(2k+1)\pi:k\in\mathbb Z\}}.$$

This domain is symmetric about zero, and $f(-x)=-f(x)$ on it. Therefore $\boxed{f\text{ is odd}}$.