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education/math/MATH1060 (trig)/Graphing.md

@@ -40,10 +40,16 @@ To find relative points to create the above graph, you can use the unit circle:
 
 If $tan(x) = \frac{sin(x)}{cos(x})$, then:
 
-| $sin(0) = 0$                              | $cos(0) = 1$                              | $tan(0) = \frac{cos(0)}{sin(0)} = \frac{0}{1} =0$                 |
+| $sin(0) = 0$                              | $cos(0) = 1$                              | $tan(0) = \frac{cos(0)}{sin(0)} = \frac{0}{1} =0$                |
-| ----------------------------------------- | ----------------------------------------- | ----------------------------------------------------------------- |
+| ----------------------------------------- | ----------------------------------------- | ---------------------------------------------------------------- |
-| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $tan(\frac{\pi}{4} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}{}}$ |
+| $sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ | $tan(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}/\frac{\sqrt{2}}{2} = 1$ |
-| $sin(\frac{\pi}{2}) = 1$                  | $cos(\frac{\pi}{2}) = 0$                  |                                                                   |
+| $sin(\frac{\pi}{2}) = 1$                  | $cos(\frac{\pi}{2}) = 0$                  | $tan(\frac{\pi}{2}) = \frac{1}{0} = DNF$                         |
+Interpreting the above table:
+- When $x = 0$, $y = 0$
+- When $x = \frac{\pi}{4}$, $y = 1$
+- When $x = \frac{\pi}{2}$, there's an asymptote
+
+Without any transformations applied, the period of $tan(x) = 1$. Because $tan$ is an odd function, $ 
 
 $$ y = cot(x) $$
 ![Graph of cotangent](assets/graphcot.svg)