I am attempting an assignment in the MIT course "Signals and Systems". There is a question I don't understand. Here it goes:
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.
a) $ \\\omega _x=\frac\pi3$, $\tau_x=0, \theta_x=2\pi,\omega _y=\frac\pi3$, $\tau_y=1, \theta_y=\frac{-\pi}{3} $
The answer:
If $\omega _x(t + \tau_x) + \theta_x=\omega _y(t + \tau_y) + \theta_y+2\pi k\\$ for any integer k, then $y(t)=x(t)$ for all $t$
Since it is satisfied for this question, the combination is right.
I dont understand how this is equal. If you sub in the values, how is $\cos(\frac\pi3t)=\sin(\frac\pi3t)$? The formula doesn't make sense to me as a result.
Am I missing something? Here is the actual question: https://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/assignments/MITRES_6_007S11_hw02.pdf
basically use the identities ... $$\sin(x) = cos(\frac\pi2 - x) =cos(x-\frac\pi2) $$ for convenience of notation I will absorb the $\tau$'s and $\theta$'s into phase shift angles $\alpha$ ...
Let $\alpha _x \equiv \omega_x\tau_x+\theta_x$
and $\alpha _y \equiv \omega_y\tau_y+\theta_y$
So we have $$ \cos( \omega_x t +\alpha_x) =\sin( \omega_y t +\alpha_y)$$ using the first identity we get $$ \cos( \omega_x t +\alpha_x) =\cos(\frac \pi 2-( \omega_y t +\alpha_y))$$
This will be satisfied when the arguments of the trig functions differ by an integer multiple of $2\pi$
the equation ... $$ \omega_x t +\alpha_x =2 k \pi +\frac \pi 2-( \omega_y t +\alpha_y)$$
must be true for all real $t$ and some integer $k$ since true for all $t$ we must have $$ \omega_x=-\omega_y \text{ and }\alpha_x+\alpha_y = (2k+\frac 12)\pi$$
using the second identity you get another set of possibilities ... $$ \omega_x=\omega_y \text{ and }\alpha_x-\alpha_y = (2k-\frac 12)\pi$$