How does one rotate a function? If I have an equation $y = \cos(x)$;
What would be the the rotated function, if I had rotated the axis by $45$ degrees ?
For example, $\cos(x)$ oscillates about the $x$ axis.
How do I determine the function which oscillates like $\cos(x)$ about the line which makes intersects the $y$ axis at $(1,0)$ and $x$ axis at $(0,1)$ ?
 A: First of all, you must parametrize your function. Let $t$ the parameter. Then:
$$ \displaystyle\left[\begin{array}{c}x \\ y\end{array} \right] = \left[\begin{array}{c}t \\ \cos(t)\end{array} \right]$$
Now, consider to rotate the function of an angle $\theta$. Then you obtain the followings:
$$ \displaystyle\left[\begin{array}{c}x' \\ y'\end{array} \right] = \left[\begin{array}{cc}\cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)\end{array} \right]\left[\begin{array}{c}t \\ \cos(t)\end{array} \right] = \left[\begin{array}{c}t\cos(\theta) + \cos(t)\sin(\theta) \\ -t\sin(\theta) + \cos(\theta)\cos(t)\end{array} \right]$$
In general, you cannot write $y'$ as a function of $x'$. This can be done in some interval. In fact, if you think to rotate a function, then you will not obtain a function, since $y'(x')$ can assume more than one value (try to figure out this by drawing).
A: You can't really speak about rotating a function. What you're describing sounds like you're rotating the graph of a function. This will sometimes but not always result in a geometrical figure that is the graph of another function.
(For example consider the function $f(x)=0$ whose graph is the $x$ axis. If you rotate that by 90° around the origin you get the $y$ axis, which is not the graph of any function).
It happens that if you turn the graph of the cosine function by 45° you will get something that is a graph for a function -- but that function doesn't have any nice formula that describes how to compute it; the best you can say is "the function whose graph arises in such-and-such way". Its actual values have to be approximated numerically.
It may be that what you really want is the function
$$ g(x) = 1-x+\cos x $$
but the graph of that is not the familiar cosine curve turned by 45°.
