Starting point of a Sine wave We learned this at school, the function:
$$y= a + b\sin (c(x-d))$$
has a starting point of $(d,a)$. But when I had to draw this function:
$$g(x) = -2-\cos(x-1/2π)$$
I thought the starting point would be $(1/2π,-2)$ but the correction model showed it to be $(1/2π,-3)$. There are multiple cases in which I run to the same problem with the starting point. Why is this? Can someone please explain? 
 A: For $\sin x:$
$$f(d)=a+b\sin(c(d-d))=a+b\sin 0=a$$
For $\cos x:$
$$g(d)=a+b\cos(c(d-d))=a+b\cos 0=a+b$$
In your case: $g\left(\dfrac{\pi}{2}\right)=-2-\cos 0=-3$
A: The difference is that $\sin(x)$ "starts at" $(0,0)$ and $\cos(x)$ "starts at" $(0,1)$ (note: By "starts at", what we really mean is "the y-intercept is at").
What you are doing is graphing sines and cosines by taking into account elementary transformations - horizontal/vertical shifts/dilations, etc. and seeing where this point ends up under these transformations.
In the case of $y = a + b\sin(c(x - d))$, we have a horizontal shift right $d$ units, a vertical dilation of $b$ units, and then a vertical shift up $a$ units. So, if we track where the y-intercept goes under these transformations, it goes from $(0,0)$ to $(d,0)$ after the horizontal shift, the dilation does nothing to this point (since $0*b = 0$!), and then finally to $(d,a)$ after we do the vertical shift.
For a $\textit{cosine}$ however, the original y-intercept is $(0,1)$, so, for $y = a + b \cos(c(x-d))$, the horizontal shift takes us to $(d,1)$, the dilation to $(d,b)$, and the vertical shift to $(d,b+a)$.
