Answer
We have $\cos\left(\theta+\dfrac{\pi}{2}\right)=-\sin\theta$ and $\sin\left(\theta+\dfrac{\pi}{2}\right)=\cos\theta$.
Work Step by Step
Consider the figure below.
We have
$\angle AOC+\angle AOB+\angle BOD=\pi$ [Angle above a straight line]
$\implies \theta+\dfrac{\pi}{2}+\angle BOD=\pi$
$\implies \angle BOD=\pi-\theta-\dfrac{\pi}{2}$
$\implies \angle BOD=\dfrac{\pi}{2}-\theta$
Now, using angle sum property for triangle in triangle $BOD$,
$\angle OBD+\angle BDO+\angle BOD=\pi$
$\implies \angle OBD+\dfrac{\pi}{2}+\dfrac{\pi}{2}-\theta=\pi$
$\implies \angle OBD+\pi-\theta=\pi$
$\implies \angle OBD=\pi-\pi+\theta$
$\implies \angle OBD=\theta$
Thus, in triangle $AOC$ and triangle $OBD$, we have $\angle AOC=\angle OBD=\theta$ and $\angle ACO=\angle BOD$, so, both right angled triangles are similar to each other.
Similar triangles have equal ratio of corresponding side lengths.
Note that both have hypotenuse of length 1 unit, so, the ratio of hypotenuse is 1 unit.
Thus, $\dfrac{BD}{OC}=\dfrac{AC}{OD}=\dfrac{AO}{BO}=1$. So, the corresponding sides are equal in length.
We have $AC=b$ and $OC=a$, so, $OD=b$ and $BD=a$ (Both $a$ and $b$ are positive). Now, using cartesian system, we get $c=-b$ and $a=d$, since, $c$ is on negative x-axis and is $b$ units away from origin.
Using definition of $\cos\theta$ and $\sin\theta$:
For point $A$, $\cos\theta=a$ and $\sin\theta=b$.
For point $B$, $\cos\left(\theta+\dfrac{\pi}{2}\right)=c$ and $\sin\left(\theta+\dfrac{\pi}{2}\right)=d$.
Thus, $\cos\left(\theta+\dfrac{\pi}{2}\right)=-\sin\theta$ and $\sin\left(\theta+\dfrac{\pi}{2}\right)=\cos\theta$.
