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?
Geometric intuition for why $\int_0^\theta \cos xdx = \sin\theta$
calculusintuitiontrigonometric-integrals
Best Answer
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.