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In the following diagram I understand how to use angle $\theta$ to find cosine and sine. However, I'm having a hard time visualizing how to arrive at tangent. Furthermore, is it true that in all right triangle trig ratios we always need to use one of the non-right angles?

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  • $\begingroup$ how did you find cosine or sine? $\endgroup$ – Nosrati Oct 8 '18 at 17:22
  • $\begingroup$ You may (or may not) find my document on this matter helpful: "(Almost) Everything You Need to Remember about Trig, in One Simple Diagram" (PDF). (It's currently undergoing revisions, so please pardon some rough spots.) $\endgroup$ – Blue Oct 8 '18 at 18:14
  • $\begingroup$ Nosrati - sin $\theta$ is just opposite over hypotenuse. Since the hypotnuse is always 1 in the unit circle sin $\theta$ will equal the height of the triangle and Y coordinate on the circle. I will now read the answers for finding tangent $\theta$ $\endgroup$ – user27343 Oct 9 '18 at 4:07
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If you pay attention, the smaller left right triangle is similar to the larger right triangle adjacent to it as they are both right triangles with angles of measure $\frac{\pi}{2}$, $\theta$, and $(\frac{\pi}{2}-\theta)$.

When two triangles are similar, the ratio of their corresponding sides will be equivalent.

$$\frac{\cos\theta}{1}$$ In the smaller right triangle, $\cos\theta$ is opposite the $(\frac{\pi}{2}-\theta)$ angle and $1$ is the hypotenuse.

In the larger right triangle, $\sin\theta$ is opposite to the $(\frac{\pi}{2}-\theta)$ angle and $x$ (unknown variable) is the hypotenuse.

$$\frac{\sin\theta}{x}$$

Using the rules of similarity, we can say the two ratios are equivalent.

$$\frac{\cos\theta}{1} = \frac{\sin\theta}{x}$$ $$x\cos\theta = \sin\theta \implies x = \frac{\sin\theta}{\cos\theta} =\implies \boxed{x = \tan\theta}$$

You can check this page out for $\csc\theta$, $\sec\theta$, and $\cot\theta$: graphical representation of trig functions.

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  • $\begingroup$ Thank you. The link was very helpful as well. $\endgroup$ – user27343 Oct 9 '18 at 4:22
  • $\begingroup$ No problem. :-) $\endgroup$ – KM101 Oct 9 '18 at 12:23
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  • In the first and third quadrants, $\tan(\theta)$ is the length from $(\cos(\theta),\sin(\theta))$ to the $x$ axis along the line tangent to the circle at $(\cos(\theta),\sin(\theta))$. In the second and fourth quadrants the situation is essentially the same but we use the opposite sign.
  • One can think of $\cos(\pi/2)$ and $\sin(\pi/2)$ from the triangle point of view as the ratios of a degenerate triangle with angles $(\pi/2,\pi/2,0)$, which is really just a line segment. Now you identify one of the $\pi/2$'s as the "right" angle and the other as the "acute" angle and measure ratios relative to that "acute" angle.
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The angle between the red and blue lines is $\theta$. So in that triangle, $$\cos\theta=\frac{\rm red}{\rm blue}$$ Or$${\rm blue}=\frac{\rm red}{\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

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  • $\begingroup$ There are a bunch of minor errors here. $\endgroup$ – Ian Oct 8 '18 at 17:24
  • $\begingroup$ Please feel free to edit the response, to make it more accurate/clear. $\endgroup$ – Andrei Oct 8 '18 at 17:25
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Commit this SohCahToa Table to memory so that you can always write it out:

enter image description here

For the $sin$ and $cos$ in the unit circle you can think of putting the values in the table over a big $1$ for the hypotenuse.

Example: Think of

$sin(\frac{\pi}{6})= \frac{{\frac{\sqrt1}{2}}}{1}$

$cos(\frac{\pi}{6})= \frac{{\frac{\sqrt3}{2}}}{1}$

and the unit circle placement of $\left(cos(\frac{\pi}{6}),sin(\frac{\pi}{6})\right)$.

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