Derivation of Wigner Function of trigonometric functions with phase The Wigner Function of $x(t)$ is
$$ W(t,f) = \frac{1}{2\pi}\int x\Big(t+\frac{\tau}{2}\Big)x^*\Big(t-\frac{\tau}{2}\Big) e^{-j2\pi f\tau}\;d\tau $$
I know how to get the $W(t,f)$ of $x(t)=\cos(2\pi f_i t)$ and $x(t)=\sin(2\pi f_i t)$ which are
$$[\delta(f+f_i)+\delta(f-f_i) + 2\delta(f) cos(4 \pi f_i t)]$$
and
$$[\delta(f+f_i)+\delta(f-f_i) - 2\delta(f) cos(4 \pi f_i t)]$$
respectively. These matched in most references for example in chapter 19 of The handbook of formulas and tables for signal processing by Poularikas.
However, I found a difficulty, also can't find proper reference, in deriving both when I add their phases $\theta$,
$x(t)=\cos(2\pi f_i t + \theta)$ and $x(t)=\sin(2\pi f_i t+ \theta)$.
 A: Given
$$\begin{align}W\left\{\cos(2\pi f_i t')\right\}(t',f)&=\dfrac{1}{2\pi}\int_{-\infty}^\infty \cos\left(2\pi f_i \left[t' + \dfrac{\tau}{2}\right]\right) \cos^*\left(2\pi f_i \left[t' - \dfrac{\tau}{2}\right]\right)e^{-j2\pi f\tau} d\tau\\
\\
& = [\delta(f+f_i)+\delta(f-f_i) + 2\delta(f) \cos(4 \pi f_i t')]\\
\end{align}$$
use the substitution
$$ t' = t + \dfrac{\theta}{2\pi f_i}$$
And write
$$\begin{align}W\left\{ \cos(2\pi f_i t')\right\}(t',f) &= \dfrac{1}{2\pi}\int_{-\infty}^\infty \cos\left(2\pi f_i \left[t +\dfrac{\theta}{2\pi f_i} + \dfrac{\tau}{2}\right]\right) \cos^*\left(2\pi f_i \left[t+\dfrac{\theta}{2\pi f_i} - \dfrac{\tau}{2}\right]\right)e^{-j2\pi f\tau} d\tau\\
\\ 
&=\left[\delta(f+f_i)+\delta(f-f_i) + 2\delta(f) \cos\left(4 \pi f_i \left[t + \dfrac{\theta}{2\pi f_i}\right]\right)\right] \\
\\
&= \dfrac{1}{2\pi}\int_{-\infty}^\infty \cos\left(2\pi f_i \left[t + \dfrac{\tau}{2}\right]+\theta\right) \cos^*\left(2\pi f_i \left[t - \dfrac{\tau}{2}\right]+\theta\right)e^{-j2\pi f\tau} d\tau \\
\\
&= W\left\{\cos(2\pi f_i t+\theta)\right\}(t,f) \\
\\
&=\left[\delta(f+f_i)+\delta(f-f_i) + 2\delta(f) \cos\left(4 \pi f_i t + 2\theta\right)\right] \\
\end{align}$$
The derivation for $\sin(2\pi f_i t +\theta)$ is analogous.
