[8.4] Inverse Trigonometric Functions
Inverse of sine
We can graph sin(x) by using parametric plotting with
$x(p) = p$ and $y(p) = \sin( p )$.
In Mathematica:
$y(p) = p$
$x(p) = \sin(p)$
But this does not pass the vertical line test, so it's not a function.
$\sin^{-1}(\sin(\theta)) = \theta$ with
domain: $[-1,+1]$
range: $[-\pi/2,\pi/2]$
So, $\sin^{-1}(1/\sqrt{2})$ can be translated as:
The angle between $-\pi/2$ and $+\pi/2$ which has a sine of $1/\sqrt{2}$
That is... $\sin^{-1}(1/\sqrt(2)) = \pi/4 = 45^o$
$\sin^{-1}(x) \equiv \arcsin(x)$
arccos
domain: $[-1,1]$
range: $[0,\pi]$
So, $\cos^{-1}(\cos((9\pi)/7))$ means
The angle between $0$ and $+\pi$ which has the same cosine as the angle $(9\pi)/7$
$\cos(\theta)$ is the $x$-coordinate of the intersection of the terminal ray with the unit circle. and has the same $x$-coordinate (pink dot) as the blue angle which is $\pi -(2\pi)/7 = (5\pi)/7$
So, $\cos^{-1}(\cos((9\pi)/7)) = (5 \pi)/7$!
$\cos^{-1}(x) \equiv \arccos(x)$
arctangent
domain: $[-\infty,+\infty]$
range: $[-\pi/2,\pi/2]$
Problem 45
- Find a formula fo the sinusoidal curve.
- Use your formula to find the $x$-coordinates of points $P$ and $Q$
1.) Express $\theta$ as the difference between two angles in the diagram which are part of right triangles. (Call them whatever you want, but label them on a sketch.)
2.) For each of your new angles, come up with an expression for the angle in terms of an arctangent. (Your expression will also involve $x$.)
3.) Put the results of these two sections together to get $\theta(x)$
4.) Graph $\theta(x)$, and show that the biggest viewing angle occurs at a distance somewhere between 5 and 6 feet from the wall.