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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?

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$1/2π$ is ambiguous. You probably mean $\pi / 2$, i.e. $\dfrac{\pi}{2}$ and $\dfrac{1}{2}\pi$, but it could be read as $\dfrac{1}{2\pi}$. – Henry Sep 11 '12 at 18:51
up vote 2 down vote accepted

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: $g\left(\dfrac{\pi}{2}\right)=-2-\cos 0=-3$

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I don't understand these answers guys. We haven't used these methods yet I believe.. – JohnPhteven Sep 11 '12 at 18:26
@ZafarS What don't you understand? $\sin 0=0$ and $\cos 0=1$, that's all there is to it. Picture the unit circle! – user39572 Sep 11 '12 at 18:27
But why do I find all over the internet that you can find the starting point by simple getting (d,a)? – JohnPhteven Sep 11 '12 at 18:29
Of course I understand the method/mathematics behind it and the reason why you can do it like this, however, I don't believe me teacher told us anything about this way and he is very strict and he'd probably wouldn't credit me for getting the right answer by using a different method – JohnPhteven Sep 11 '12 at 18:33
Oh wait a sec I've been incredibly stupid I'm extremely sorry! Thank you Julien! How do I give you like a 'best answer' or doesn't this site work like that :P? – JohnPhteven Sep 11 '12 at 18:37

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 $(d,b+a)$.

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