Grammar fixes

public/content/lessons/2017/derivatives-trig-functions/index.mdx

@@ -14,7 +14,7 @@ credits:
 
 Let's try to reason through what the derivatives of the functions sine and cosine should be. For background, you should be comfortable with how to think about both of these functions using the unit circle; that is, the circle with radius $1$ centered at the origin[^1].
 
-[^1]: To get up to speed on the unit circle or for a quick referesher see [this article](https://wumbo.net/concepts/unit-circle/).
+[^1]: To get up to speed on the unit circle or for a quick refresher see [this article](https://wumbo.net/concepts/unit-circle/).
 
 For example, how would you interpret the value $\sin(0.8)$ if the value $\theta = 0.8$ is understood to be in radians?
 You might imagine walking around a circle with a radius of $1$, starting from the rightmost point, until you’ve traversed the distance $0.8$ in arc length.
@@ -33,15 +33,15 @@ So the graph of $\sin(\theta)$ vs. $\theta$, which plots this height as a functi
   video="figures/13.22-13.38.mp4"
 />
 
-Just from looking at this graph, we can get a feel from the shape of the derivative function. The slope at $0$ is something positive, then as $\sin(\theta)$ approaches its peak, the slope goes down to $0$. Then the slope is negative for a little while before coming back up to $0$ as the $\sin(\theta)$ graph levels out. If you’re familiar with the graphs of trig functions, you might guess that this derivative graph should be exactly $\cos(\theta)$, whose graph is just a shifted-back copy of the sine graph.
+Just from looking at this graph, we can get a feel for the shape of the derivative function. The slope at $0$ is something positive, then as $\sin(\theta)$ approaches its peak, the slope goes down to $0$. Then the slope is negative for a little while before coming back up to $0$ as the $\sin(\theta)$ graph levels out. If you’re familiar with the graphs of trig functions, you might guess that this derivative graph should be exactly $\cos(\theta)$, whose graph is just a shifted-back copy of the sine graph.
 
 <Figure
   image="figures/13.44-cosine.svg"
   video="figures/13.40-14.30.mp4"
   show="video"
 />
 
-But all this tells us is that the peaks and valleys of the derivative graph seem to line up with the graph of cosine. How could we know that this derivative actually *is* the cosine of theta, and not just some new function that happens looks similar to it. As with the previous examples of this video, a more exact understanding of the derivative requires looking at what the function itself represents, rather than the graph of the function.
+But all this tells us is that the peaks and valleys of the derivative graph seem to line up with the graph of cosine. How could we know that this derivative actually *is* the cosine of theta, and not just some new function that looks similar to it? As with the previous examples of this video, a more exact understanding of the derivative requires looking at what the function itself represents, rather than the graph of the function.
 
 Think back to the walk around the unit circle, having traversed an arc length of $\theta$, where $\sin(\theta)$ is the height of this point. Consider a slight nudge of $d$-theta along the circumference of the circle; a tiny step in your walk around the unit circle. How much does this change $\sin(\theta)$? How much does that step change your *height* above the x-axis? This is best observed by zooming in on the point where you are on the circle.
 
@@ -57,15 +57,15 @@ Consider the right triangle pictured below, where the hypotenuse represents a st
 />
 
 This tiny triangle is actually similar to this larger triangle with the defining angle theta, and whose hypotenuse is the radius of the circle with length $1$.
-Specifically, the angle between its height $d(\sin(\theta))$ and its hypotenuse $d\theta$ is precisely equals to $\theta$.
+Specifically, the angle between its height $d(\sin(\theta))$ and its hypotenuse $d\theta$ is precisely equal to $\theta$.
 
 <Figure
   image="figures/similar-triangles.svg" 
 />
 
 <Accordion title="Similar triangle explanation">
 
-We know these right triangles must be similar by doing some angle chasing geometry. Start by drawing a line parallel to the $x$-axis. From geometry, we know that the angle formed between the radius and this parallel line is equal to theta. 
+We know these right triangles must be similar by doing some angle-chasing geometry. Start by drawing a line parallel to the $x$-axis. From geometry, we know that the angle formed between the radius and this parallel line is equal to theta. 
 
 <Figure
   image="figures/similar-triangles-line-parallel-to-x-axis.svg" 
@@ -87,19 +87,19 @@ Since these two triangles share the same angles, they must be similar triangles,
 
 </Accordion>
 
-Think about what the derivative of sine is supposed to mean. It’s the ratio between that $d\left(\sin(\theta)\right)$, the tiny change to the output of sine, divided by $d \theta$, the tiny change to the input of the function. From the picture, that’s the ratio between the length of the side adjacent to this little right-triangle divided by the hypotenuse. Well, let’s see, adjacent divided by hypotenuse; that’s exactly what $\cos(\theta)$ means!
+Think about what the derivative of sine is supposed to mean. It’s the ratio between that $d\left(\sin(\theta)\right)$, the tiny change to the output of sine, divided by $d \theta$, the tiny change to the input of the function. From the picture, that’s the ratio between the length of the side adjacent to this little right triangle divided by the hypotenuse. Well, let’s see, adjacent divided by hypotenuse; that’s exactly what $\cos(\theta)$ means!
 
 <Figure
   image="figures/16.01.svg" 
 />
 
 Notice, by considering the slope of the graph, we can get a quick intuitive feel for the rough shape that the derivative of $\sin(\theta)$ should have, which is enough to make an educated guess. But to more to understand why this derivative is *precisely* $\cos(\theta)$, we had to begin our line of reasoning with the defining features of $\sin(\theta)$.
 
-For those of you who enjoy pausing and pondering, take a moment to find a similar line of reasoning which explains what the derivative of $\cos(\theta)$ should be.
+For those of you who enjoy pausing and pondering, take a moment to find a similar line of reasoning that explains what the derivative of $\cos(\theta)$ should be.
 
 <Accordion title="Derivative of Cosine">
 
-To find the derivative of the cosine function, let's look at the unit circle definition of cosine. Imagine walking around the unit circle starting from the right-most point an arc-length distance of $\theta$ (theta) unit. The cosine function returns the distance from the $y$-axis and the end-point corresponding the the angle.
+To find the derivative of the cosine function, let's look at the unit circle definition of cosine. Imagine walking around the unit circle starting from the right-most point to an arc-length distance of $\theta$ (theta) unit. The cosine function returns the distance from the $y$-axis and the end-point corresponding the angle.
 
 <Figure
   image="figures/cosine-unit-circle.svg"