Why is Mathematica converting Sin(x + pi/2) to Cos(x)?

Question

My grandson and I are trying to plot Sin[x] and Sin[x + pi/2] on the same axis.

Sin[x + pi/2] should be similar in magnitude and frequency to the Sin[x] curve, but shifted pi/2 to the left. The problem is that Mathematica is converting Sin[x + pi/2] to Cos[x]. When we try to plot these together, we get the following:

As you can see, the Sin[x + pi/2] (now Cos[x]!) represented by the light-brown curve is centered at the y-axis, instead of being shifted pi/2 to the left. Also, the Sin[x] curve has been shifted to the right instead of being centered on the y-axis.

Why is this happening? Why is Mathematica converting Sin[x + Pi/2] to Cos[x]? Also, wouldn't you expect the Sin[x] curve (in blue) to also be centered on the y-axis?

Here's our code:

 y1[x_] := Sin[x];
 y2[x_] := Sin[x + Pi/2];
 a = -2 Pi; b = 2 Pi;
 Plot[{y1[x], y2[x]}, {x, a, b}]

Instead of Pi we have the symbol for pi in our actual code.

Answer 1

The reason why Sin[x+Pi/2] is converted to Cos[x] is, that it is the simplest form. This is the way Mathematica works. You input an expression and Mathematica tries to normalize it as much as possible by applying rules that are coded into the system. There are many many rules and more importantly, often you wouldn't recognize them as transformations of expressions. What about this

Plus[1, 1]
(* 2 *)

I hope you agree that you wouldn't complain about this transformation. In your case, it is exactly the same although it is not as obvious as 1+1. Cos[x] is just the best form Mathematica could find after applying the rules of the system.

Also, wouldn't you expect the Sin[x] curve (in blue ) to also be centered on the y-axis?

That is a question I don't understand, but Sin[x] just looks this way. Maybe you can clarify this a bit.

Answer 2

Sin[x + Pi/2] can be written in an easier way due to mathematical formula:

$\sin(a + b) = \sin(a)\cos(b) + \sin(b)\cos(a)$

Here, $a = x$ and $b = \pi/2$. You need to know that $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$.

So you rewrite with the formula:

$\sin(x + \pi/2) = \sin(x)\cos(\pi/2) + \sin(\pi/2)\cos(x)$

$= \sin(x)\cdot 0 + 1\cdot \cos(x)$

$= \cos(x)$

Mathematica just uses a simpler form but both expressions are exactly the same.

Answer 3

I don't think the issue here is Mathematica at all; rather, I think you are confused about what the graph of $y=\sin x$ is supposed to look like.

The function $y=\sin x$ is not "centered on the $y$-axis"; rather, it has odd symmetry, i.e. $180^\circ$ rotational symmetry around the origin. $y=\sin x$ is shown below:

The graph of $y=\sin (x + \pi/2)$ is the same as $y=\sin x$ but shifted $\pi/2$ units (i.e. one-quarter period) to the left, which has the effect of moving the maximum to the $y$-axis:

This function, unlike the "unshifted" version, is symmetric across the $y$-axis. And it also happens to be completely identical with the function $y=\cos x$, which has even symmetry.

So now go back to the original graph you included in your post. The blue curve, $y=\sin x$, has not "been shifted to the right instead of being centered on the y-axis". It's right where it's supposed to be, and should not be centered on the $y$-axis. When you do shift it to the left, then it ends up centered on the $y$-axis, and exactly equal to the cosine function.