Section+1.9

Today in class, we discussed the sine, cosine, and tangent functions and their inverses. Here are the notes:


 * In order to be a function, a graph must pass the vertical line test (all vertical lines cross the graph at most once)
 * In order for a function to have an inverse, it must also pass the horizontal line test (all horizontal lines cross the graph at most once)
 * A function that passes both the horizontal and vertical line tests is called a one to one function.
 * Algebraically, the inverse of a function undoes the function, so you solve for x(independent value), and then **switch x and y**
 * Graphically, the inverse function is the original graph reflected over the line y=x. Notice that this **switches x and y** in the coordinates.
 * Domain of inverse=Range of original; Range of inverse=Domain of original

Minds in Action:

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When (as Sasha and Tony did) we took the $\tan ^{-1}$ of 3 angles, we discovered that they were all in between -1.57 and 1.57. This is because $1.57=\frac{\pi }{2}$, which is the period of tangent. The calculator only gives values in between $\frac{\pi }{2}$ and $\frac{\pi }{2}$ because within the restricted domain of $(\frac{-\pi }{2}, \frac{\pi }{2})$, tangent has an inverse function.

The next thing that we did in class was graph tangent and $\tan ^{-1}$, as well as $\sin ^{-1}$ and $\cosn ^{-1}$. Here are those graphs:

Tangent and $\tan ^{-1}$:

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The red line is tangent of x on the restricted domain of $(\frac{-\pi }{2}, \frac{\pi }{2})$, and the green line is $\tan ^{-1}$ on the same, restricted domain. On this domain, the domain of tangent is $(\frac{-\pi }{2}, \frac{\pi }{2})$, and the range is $\mathbb{R}$. Likewise, the domain for $\tan ^{-1}$ is $\mathbb{R}$, and the range is $(\frac{-\pi }{2}, \frac{\pi }{2})$

Sin(x) and $\sin ^{-1}$:

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The restricted domain of $\sin (x)$ is $[-\frac{\pi }{2}, \frac{\pi }{2}]$ Range: [-1, 1]

The domain of $\sin^{-1}$ is [-1, 1], and the range is $[-\frac{\pi }{2}, \frac{\pi }{2}]$

Cos(x) and $\cos ^{-1}$:

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