When you let t = 0 in [itex]6\cos({\omega}t + \varphi)[/itex], you get [itex]6\cos\varphi = 6[/itex], giving [itex]\varphi=0[/itex]. So the equation should simply be [itex]6\cos{\omega}t[/itex].

This is not one of the answers. However, you should be able to see (from basic trig) that [itex]\sin(\theta + \frac{\pi}{2}) = \cos\theta[/itex]. So you can match it to the first answer.

It doesn't matter if you use sine or cosine, both are equally valid. They may look different, but if you do the basic math, they're actually identical.

And here, you don't actually have to derive very much. It's easier to just go through the multiple choices to see which fits, i.e. which gives a displacement of 6 when t = 0.