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Can someone please explain to me why cotangent graphs look the way they do? I want to know why they basically look like mirror reflected tangent graphs. I get that if $\tan\theta=y/x$, then $\cot\theta=x/y$. But why would this lead to a graph that looks like a tangent graph that was reflected over?

Can you please try to keep the answers at the level of a high school pre-calc student?

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    $\begingroup$ Not sure how helpful this is (especially because of how it's inverted) but the cotangent function can be thought of as recording the x value of the intersection of the line $y=1$ and the line that travels through origin to a point on the unit circle. Here is a visual of this: desmos.com/calculator/qbdj61u2t9 $\endgroup$ – Mason Aug 3 '18 at 4:16
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We have $$ \cot(x) = \frac{\cos(x)}{\sin(x)} = \frac{\sin(\pi/2-x)}{\cos(\pi/2-x)} = \tan(\pi/2-x)$$

so it is reflected and shifted by $\pi/2.$

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  • $\begingroup$ But how do we know this till holds when theta, or x in this case, is greater than pi/2? For instance, what if x was 3pi/4? How would this still hold? Can you prove it me please? $\endgroup$ – Ethan Chan Aug 3 '18 at 4:07
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    $\begingroup$ @EthanChan Do you know what the graphs of $\sin$ and $\cos$ look like for all $x$? $\endgroup$ – spaceisdarkgreen Aug 3 '18 at 4:17
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    $\begingroup$ Note: "it is reflected and shifted by $π/2$" holds for every trigonometric function when you change function to cofunction or the other way around. Quite useful. The answer uses this formula with $cos$ and $sin$ to obtain it for $cot$. $\endgroup$ – Kamil Maciorowski Aug 3 '18 at 6:50
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    $\begingroup$ @EthanChan: In case you were taught the definition of $\sin,\cos$ in terms of the right-angle triangle, here is the correct way to extend it to other angles. Consider the unit circle $C$ in the $x,y$-plane (centred at origin with radius $1$). For angle $t$, draw a ray $R$ through the origin that is the result of rotating the $x$-axis by $t$ anti-clockwise. Let $P$ be the intersection of $R$ and $C$. Then define $\cos(x),\sin(x)$ as the $x,y$ coordinates of $P$. Equivalently, $P$ is the result of starting from $(1,0)$ and moving anti-clockwise along $C$ by a distance (arc-length) of $t$. $\endgroup$ – user21820 Aug 3 '18 at 11:59
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    $\begingroup$ @EthanChan: Then observe that if you change the angle from $t$ to $π/2-t$ it is equivalent to reflecting $P$ across the line $x=y$. This is the geometric reason why $\cos(t) = \sin(π/2-t)$ and $\sin(t) = \cos(π/2-t)$, because the $x,y$-coordinates are swapped in such a reflection. $\endgroup$ – user21820 Aug 3 '18 at 12:02
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Just as a visual complement for the answers you already have, notice the following animation.

Green angle: $\theta$, yellow angle: $\pi/2 - \theta$ and $\cot(x)$ is the red curve.

enter image description here

For more details regarding these types of constructions, see this question.

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$$\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\sin(\frac\pi2-\theta)}{\cos(\frac\pi2-\theta)} = \tan\Big(\frac\pi2-\theta\Big)$$

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  • $\begingroup$ Okay. Last thing, how does this work for values of theta greater than 90 degrees? $\endgroup$ – Ethan Chan Aug 3 '18 at 3:58
  • $\begingroup$ These identities are always true, as long as $\sin\theta\neq0$. $\endgroup$ – mr_e_man Aug 3 '18 at 3:59
  • $\begingroup$ But can you prove why they will always be true? $\endgroup$ – Ethan Chan Aug 3 '18 at 4:08
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    $\begingroup$ That depends on your definitions of the trig functions. Cotangent is always cosine over sine, and tangent is always sine over cosine. All that remains is whether $\cos\theta = \sin(\pi/2-\theta)$ (which is equivalent to the same with $\cos$ and $\sin$ swapped; try $\theta=\pi/2-\phi$). $\endgroup$ – mr_e_man Aug 3 '18 at 4:10
  • $\begingroup$ If you're using the unit circle, defining $\cos\theta=x$ and $\sin\theta=y$, then swapping $\sin\theta$ and $\cos\theta$ is the same as reflecting across the line $y=x$. This reflects the angle relative to $\pi/4$. $$\theta'-\frac\pi4=-\Big(\theta-\frac\pi4\Big)$$ $$\theta'=\frac\pi2-\theta$$ $$\cos\theta=\sin\theta'=\sin(\pi/2-\theta)$$ $\endgroup$ – mr_e_man Aug 3 '18 at 4:23

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