trigonometry
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?
Ok, I have found how to fit my sine curve. I have done it by adding delta y between those two vectors into the value of found sine function (blue). However, the sine wave which is now fitted between points is a little bit different (violet). I got a question for all you experts, does it affect the length of sine wave which i found or the length of the original and "moved" curves are the same? Here is the image
It's because clockwise angles are taken to be negative while counterclockwise angles are positive. Thus to define the functions, we work on positive angles, hence counterclockwise.
You may ask, why are counterclockwise angles positive and not the other way round?
The answer lies in our choice of complex plane.
Our choice while defining complex plane was-
As a result, multiplication by $e^{i\theta}$ rotates the complex plane by the angle $\theta$ counterclockwise.
So, to avoid negative signs in our problems, we say that multiplication by $e^{i\theta}$ rotates the plane by angle $\theta$, not $-\theta$, in other words, we adopt the counterclockwise direction as positive.
Best Answer
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: $g\left(\dfrac{\pi}{2}\right)=-2-\cos 0=-3$