# Understanding sine, cosine, and tangent in the unit circle 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?

• how did you find cosine or sine? – Nosrati Oct 8 '18 at 17:22
• 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.) – Blue Oct 8 '18 at 18:14
• 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 $\theta$ – user27343 Oct 9 '18 at 4:07

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.

• Thank you. The link was very helpful as well. – user27343 Oct 9 '18 at 4:22
• No problem. :-) – KM101 Oct 9 '18 at 12:23
• 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.

The angle between the red and blue lines is $$\theta$$. So in that triangle, $$\cos\theta=\frac{\rm red}{\rm blue}$$ Or$${\rm blue}=\frac{\rm red}{\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

• There are a bunch of minor errors here. – Ian Oct 8 '18 at 17:24
• Please feel free to edit the response, to make it more accurate/clear. – Andrei Oct 8 '18 at 17:25

Commit this SohCahToa Table to memory so that you can always write it out: 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 $$\left(cos(\frac{\pi}{6}),sin(\frac{\pi}{6})\right)$$.