Derivation of $\sin(90^\circ+\theta)$, $\cos(90^\circ+\theta)$, etc How are the trigonometric functions 
like $\sin(90^\circ+\theta)$, $\cos(90^\circ+\theta)$, $\tan(90^\circ+\theta)$, $\cot(90^\circ+\theta)$, $\sec(90^\circ+\theta)$, $\csc(90^\circ+\theta)$ derived?
 A: [I looked for questions like this but so far have not found one that is really the same question, although the one cited in the comments under this question is certainly relevant. So instead of attempting to link to a duplicate, I'm answering.]
If you use the unit circle definition, to make things simple it's better for all angles to be expressed in radians, so $90$ is just $\frac\pi2.$
Then for a positive angle $\theta$ you can simply start at the point $(1,0)$ on the unit circle (on the positive $x$ axis) and travel counterclockwise along the circle until the total distance you have traveled along the arc of the circle is $\theta.$
Then $\sin(\theta)$ is the $y$ coordinate of the point you reached at the end of that path, and $\cos(\theta)$ is the $x$ coordinate of that same point.
Now let's try to find the sine of $\left(\frac\pi2 + \theta\right)$ radians,
that is, the sine of $90$ degrees plus $\theta$ radians.
One way to do this is, first we travel a distance $\frac\pi2$ counterclockwise from the point $(x,y)=(1,0).$ That gets us to the point $(x,y)=(0,1).$
Then we travel an additional distance $\theta$ from that point.
Let's give a name to the point at the end of that arc where we stop; 
call it $P$.
Now let's lay down some alternative coordinates on the plane:
two new axes that go through the same origin as the $x$ and $y$ axes.
The $x'$ axis is identical to the $y$ axis.
The $y'$ axis is concurrent with the $x$ axis, but goes in the opposite direction;
that is, points with positive $x$ coordinates have negative $y'$ coordinates.
In fact, if $y'_P$ is the $y'$-coordinate and $x_P$ is the $x$-coordinate of some point $P,$ then $y'_P = -x_P.$
The point with coordinates $(x,y)=(0,1)$ in the original coordinate system has coordinates $(x',y')=(1,0)$ in the new system.
In fact, the figure composed of the $x'$ axis, $y'$ axis, and the arc of length $\theta$ from
$(x',y')=(1,0)$ to $P$ is exactly congruent to the figure composed of the $x$ axis, $y$ axis, and the arc of length $\theta$ from $(x,y)=(1,0)$ a point $Q$ at which we would measure $\sin(\theta).$
So just as $\sin(\theta)$ is the $y$-coordinate of $Q,$
$\sin(\theta)$ is also the $y'$-coordinate of $P.$
That is, $\sin(\theta) = y'_P.$
But remember that $P$ is the point we reach by traveling 
$\left(\frac\pi2 + \theta\right)$ radians along an arc starting at $(x,y)=(1,0).$
So $x_P,$ the $x$ coordinate of $P$, is the cosine of 
$\left(\frac\pi2 + \theta\right)$ radians.
Also remember that $y'_P = -x_P,$ or equivalently $x_P = -y_P.$
Putting these facts together, we have
$$
\cos\left(\frac\pi2 + \theta\right) = -y_P = -\sin(\theta).
$$
Looking at the $y$ and $x'$ coordinates of $P,$ using the same facts about sines and cosines of the unit circle and about the congruence of the figures,
we can show that
$$
\sin\left(\frac\pi2 + \theta\right) = y_P = x'_P = \cos(\theta).
$$
All the other trigonometric functions can be derived from these.

The related question which one also ought to be able to answer is why we accept the unit circle definition as a good definition for sine and cosine.
That question is better addressed by 
this answer,
which shows some ways in which the formulas we get by using the unit circle are consistent with the ones we have for angles between zero and a right angle.
A: Write $f(\theta)=(90+ \theta)$
Then for $g(\theta)$ any trigonometric function, your question is to find $(g(f(\theta)))'$.
By the chain rule we have $(g(f(\theta)))'=g'(f(\theta))\cdot f'(\theta)=g'(f(\theta))$, since $f'(\theta)=1$.
As an example $(\sin(90+\theta))'=\cos(90+\theta)$
