Where should triangle, sawtooth and square waves start?
If you look at it on the surface, all periodic waves have phases associated with them such as $\sin(\omega t + \phi)$ or $\cos(2\pi f_0 t + \theta)$ which has phase angels of $\phi$ and $\theta$ radians respectively. Therefore, they do not necessarily begin at their peaks at $t=0$ always. Indeed at $t=0$ above waves assume the values of $\sin(\phi)$ and $\cos(\theta)$. And you can therefore shift them by amounts of $\phi$ and $\theta$ left or right to make their phases zero with respect to display graphics plotting systems. 

Oscilloscopes make this adjustment by their trigger levels. For example, to set the peak of the signal $A \cos(2\pi f_0 t + \theta)$ at the origin ($t=0$) then you shall adjust your trigger level to about $A$, assuming that the trigger operation will start plotting the wave at that amplitude positioned at the origin of the coordinate grid.

The same should be true for other waveshapes such as sawtooth, triangle, square or many unnamed periodic possibilitied.

I cannot see if you have a deeper question here, but on the surface the the answer is as above.

Plot of $\text{sgn}( \sin( 2\pi t ) ) $ :



And plot of $\text{sgn}( \cos( 2\pi t ) ) $ :