Why $x(t) = \cos(\frac{\pi}{3}t)$ and $y(t) = \sin(\frac{\pi}{3}t)$ are considered identical signals? Consider the following question that I toke from Signals and systems: Part I exercises, question 1:


  (b) With $x(t) = \cos(\omega_x(t + \tau_x) + \theta_x)$ and
  $y(t) = sin(\omega_y(t + \tau_y)+ \theta_y)$, determine for which of the
  following combinations $x(t)$ and $y(t)$ are identically equal for all
  $t$.


Given that:
$$
\begin{matrix}
\omega_x &amp; \tau_x &amp; \theta_x &amp; \omega_y &amp; \tau_y &amp; \theta_y \\
\hline
\pi/3    &amp; 0      &amp; 2\pi     &amp; \pi/3    &amp; 1      &amp; -\pi/3 
\end{matrix}
$$

The solutions manual shows that $x(t)$ and $y(t)$ are identical with these values:


  $\omega_x = \omega_y$, $\omega_x\tau_x + \theta_x = 2\pi$, $\omega_y\tau_y + \theta_y = \frac{\pi}{3}(1) - \frac{\pi}{3} = 0 + 2\pi k$
  
  Thus, $x(t) = y(t)$ for all $t$.


It toke me sometime to understand this, but still I'm not sure how these signals are identical.

If we substitute the constants we will get:

$$
\begin{align*}
x(t) &amp;= \cos(\frac{\pi}{3}(t + 0) + 2\pi) \\
     &amp;= \cos(\frac{\pi}{3} t + 2\pi) \\
     &amp;= \cos(\frac{\pi}{3}t) \\
\\
y(t) &amp;= \sin(\frac{\pi}{3}(t + 1) - \frac{\pi}{3}) \\
     &amp;= \sin(\frac{\pi}{3}t + \frac{\pi}{3} - \frac{\pi}{3}) \\
     &amp;= \sin(\frac{\pi}{3}t)
\end{align*}
$$

which is clear that $x(t) = \cos(\frac{\pi}{3}t)$ and $y(t) = \sin(\frac{\pi}{3}t)$ are not identical:



Can someone please explain what I'm missing?
There is obviously an error in the problem formulation. From the given solution, the problem should read


  With $x(t) = \cos(\omega_x(t + \tau_x) + \theta_x)$ and
  $y(t) = \cos(\omega_y(t + \tau_y)+ \theta_y)$, determine for which of the
  following combinations $x(t)$ and $y(t)$ are identically equal for all
  $t$.


In this case the given solution $\omega_x=\omega_y$ and $\omega_x\tau_x+\theta_x=\omega_y\tau_y+\theta_y+2\pi k$ makes sense. The solution to the problem as it is given (with $y(t)=\sin(\omega_y(t + \tau_y)+ \theta_y)$), would be

$$\omega_x\tau_x+\theta_x=\omega_y\tau_y+\theta_y-\frac{\pi}{2}+2\pi k$$