Phase response of $H(f)=e^{-j2{\pi}ft_0}$

Question

Given is the impulse response: $$h(t)=\delta(t-t_0)$$.

I calculated $$H(f)=e^{-j2 \pi ft_0}=|H(f)|\cdot e^{j\varphi(f)}$$.

Now, the magnitude response of $H(f)$ is:

$$\begin{align} |H(f)| &=\sqrt{\Re\{H(f)\}^2+\Im\{H(f)\}^2} \\ &= \sqrt{\cos^2(2{\pi}ft_0)+\sin^2(2{\pi}ft_0)} \\ &= 1 \\ \end{align}$$.

Using: $$e^{-j \theta}=\cos(\theta)- j \cdot \sin(\theta)$$ and $$\cos^2(\theta)+\sin^2(\theta) = 1$$.

But now I wonder about the solution of the phase response. I know that the solution is:

$$\varphi(f)=-2 \pi f t_0$$.

I would be thankful if someone could show me how the phase response was calculated, since following

$$\varphi(f) = \arctan \left(\frac{\Im\{H(f)\}}{\Re\{H(f)\}} \right) = \arctan \left(\frac{-\sin(2 \pi f t_0)}{\cos(2 \pi f t_0)} \right)$$

I instead obtain $$\varphi(f)=-\arctan\big(\tan(2{\pi}ft_0)\big)$$ which differs from the given sample solution.

Answer 1

Wouldn't it be much easier to just compare the left-hand side and the right-hand side of

$$e^{-j2\pi ft_0}=|H(f)|e^{j\phi(f)}\tag{1}$$

to see that $|H(f)|=1$ and $\phi(f)=-2\pi ft_0$?

Answer 2

$\varphi(f)=-\arctan(\tan(2{\pi}ft_0)) = -2{\pi}ft_0$

Which is the solution