Why $x(t) = \cos(\frac{\pi}{3}t)$ and $y(t) = \sin(\frac{\pi}{3}t)$ are considered identical signals?

Question

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

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:

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

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?

Answer 1

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

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