[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:

To graph the inverse relation (in gray) of the sine function, we just swap $x(p)$ and $y(p)$:

$y(p) = p$
$x(p) = \sin(p)$

But this does not pass the vertical line test, so it's not a function.
But, restrict the angle $\theta \in [-\pi/2,\pi/2]$ and we get a function $\arcsin(x) \equiv \sin^{-1}(x)$ that behaves like....

$\sin^{-1}(\sin(\theta)) = \theta$ 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

$\arccos(x) \equiv \cos^{-1}(x)$ with
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 $(9\pi)/7$

The green angle is $(9\pi)/7 = \pi + (2\pi)/7$.

$\cos(\theta)$ 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

$\arctan(x) \equiv \tan^{-1}(x)$ with
domain: $[-\infty,+\infty]$
range: $[-\pi/2,\pi/2]$

Problem 45


  1. Find a formula fo the sinusoidal curve.
  2. 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.