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Sine Cosine Plot

Taking the above plot I'm looking for some intuition how to think about the integral of $sin(x)$, which is $-cos(x)$ (plus some constant that's assumed to be zero for the sake of readbility). The derivative of $sin(x)$ is easily interpreted visually, the slope at $\pi/2$ at the top of the $sin$ curve is zero so $cos(\pi/2)$ shows just that.

What gets my brain in a knot is why $-cos(\pi/2)$ is also zero. Maybe the hole I'm in starts with thinking about the integral as the area under the $sin$ function. This is clearly not zero, neither from 0 to $\pi/2$ nor 'at' $\pi/2$ for any non-zero slivers of x.

I'm probably not thinking about this the right way so the question boils down to what $-cos(x)$ is telling us about $sin(x)$, visually I can't make any sense of it.

I do appreciate the irony that the derivative of $-cos(x)$ is $sin(x)$!

Michael H
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2 Answers2

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Since $\int_0^x\sin t dt=1-\cos x$, all that $\cos(\pi/2) =0$ means in area terms is that $\int_0^{\pi/2}\sin t dt=1$, or equivalently $\int_{\pi/2}^x\sin t dt=-\cos x$.

J.G.
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  • Can you please take a step back and explain where the 1 in $1 - cos x$ comes from? Why does $-cos x$ not give the area directly? I thought that is what integrals do. – Michael H Sep 20 '20 at 15:22
  • @MichaelH Well, what (definite) integrals actually do is$$\int_a^x(-\cos t)^\prime dt=(-\cos x)-(-\cos a)=\cos a-\cos x.$$This simplifies with $a=0$ to $\int_a^x\sin t dt=1-\cos x$. The upshot of it is one antiderivative's roots don't tell you much, because where one antiderivative is $0$ the others are arbitrary other values. On the other hand, if $\cos a=\cos b$ then $\int_a^b\sin tdt=0$, which means areas above the $x$-axis cancel those below it. – J.G. Sep 20 '20 at 15:28
  • Ok thanks I get the 1 now and I guess I need to relearn a few things. So then definite integrals give an area, that's a start. – Michael H Sep 21 '20 at 13:52
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"Why $-\cos (\pi/2)$ is also $0$?"

This question reduces to : why $~\cos (\pi/2)$ is also $0$?

For such a question, don't think of the sine and cosine functions as represented by integrals. Instead, imagine traveling counter clockwise around the arc of the unit circle, starting at $(1,0).$

At any point along this arc, associate the (dimensionless) measure of the arc as $2\pi \times$ the proportion of a complete revolution.

Further, given a specific point $\theta$ along the arc, its $(x,y)$ coordinates are $(\cos \theta, \sin \theta).$ Now, imagine the specific point along the arc that is 1/4 of the way around, which corresponds to $1/4 \times 2\pi = \pi/2$.

The $(x,y)$ coordinate of the corresponding point is $(0, 1)$. Therefore, $\cos (\pi/2) = 0.$

user2661923
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