Geometric intuition for why $\int_0^\theta \cos xdx = \sin\theta$ Why is $\int_0^\theta \cos x dx = \sin\theta$? How does the area under the cosine curve from 0 to some angle $\theta$ relate to the unit circle definition of $\sin\theta$, i.e, as the vertical distance travelled across the unit circle circumference from 0 to $\theta$. Is there a way to see how these two geometric pictures relate to each other, so as to develop an intuition for the equality?
 A: Imagine a particle moving counterclockwise around the unit circle at unit angular velocity. Then at time $t$ the particle is at point $(\cos t, \sin t)$. The velocity vector at that point is $(-\sin t, \cos t)$. The second component, $\cos t$, is the rate at which the $y$ coordinate in increasing.
Now remember that integration is good for more things than finding areas.
The integral of velocity is the distance covered(1): the distance covered is the area under the velocity curve.
In this case the vertical distance covered in the time interval $[0, \theta]$ is $\sin \theta$, the $y$ coordinate of the point you reach. So
$$
\int_0^\theta \cos t dt  = \sin \theta.
$$
(1) That sentence is really the essence of the fundamental theorem of calculus. Remember it when you get to that theorem.
A: We can develop an intuition for the equality noting that the value of $A(\theta)=\int_0^\theta \cos xdx$ corresponds to the area under the graph for $\cos x$ between $0$ and $\theta$ and we have that

*

*for $\theta =0$ we have $A(0)=0$ and indeed $\sin 0 =0$

*for $\theta \in(0,\pi/2)$ we have that $A(\theta)$ increases and indeed also $\sin \theta$ increases

*for $\theta =\pi/2$ we have $A(\pi/2)$ reaches its maximum and indeed $\sin (\pi/2) =1$ which is its maximum

*$\ldots$
and so on according to the following sketch

