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zleyyij 2024-10-07 13:48:49 -06:00
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@ -125,6 +125,8 @@ The inverse of a trig function is **not** the same as the reciprocal of a trig f
- To find the inverse of cos, you need to restrict the domain to $[0, \pi]$
- To find the inverse of tangent, you need to restrict the domain to $(-\frac{\pi}{2}, \frac{\pi}{2})$.
The graphs of an inverse function can be found by taking the graph of $f$, and flipping it over the line $y=x$.
# Examples
> Given $-2\tan(\pi*x + \pi) - 1$
@ -145,3 +147,7 @@ Vertical shift: $1$
| Period | $\frac{\pi}{\|\pi\|} = 1$ |
| Phase shift | $\frac{-\pi}{\pi} = -1$ |
| Vertical shift | $-1$ |
> Evaluate $\arccos{\frac{1}{2}}$ using the unit circle.
Taking the inverse of the above function, we get this. Because the domain of $cos$ ranges from $0$ to $\pi$ inclusive, the answer is going to be in quadrant 1 or quadrant 2.
$$ cos(x) = \frac{1}{2} $$