Inv Trig Function BreakGround

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Two young mathematicians think about trigonometric functions.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, do you remember talking about trig equations a while ago?
Riley: Absolutely! We used trig identities to simplify the equation, then built triangles to find reference angles.
Devyn: Right., but if I have a trig equation like $\displaystyle \sin (x) = \frac {2}{5}$ ....
Riley: SOH-CAH-TOA, so sine is opposite over hypotenuse. We build a triangle whose opposite side has length $2$ , and whose hypotenuse is $5$ . The other side is given by $\sqrt {25-2} = \sqrt {23}$ .
Devyn: Yes, but what is the angle! That’s not one of the triangles whose angles I remember.
Riley: Oh no! What do we do?

Recall that two functions $f$ and $g$ are inverses of one another if both:

$f\left ( g(x) \right ) = x$ for all $x$ in the domain of $g$ .
$g\left ( f(x) \right ) = x$ for all $x$ in the domain of $f$ .

What condition do we need in order for a function to have an inverse?

It must pass the vertical line test. It must pass the horizontal line test. It must be an even function. It must be an odd function. It must be one-to-one.

Does the function $\sin (x)$ have an inverse?

No Yes

Is $\sin ^{-1}(x)$ the inverse of $\sin (x)$ ?

No Yes

Find the solution of $\sin (x) = \frac {2}{5}$ that lies in the interval $\left (0, \frac {\pi }{2}\right )$ . $x = \answer { \sin ^{-1}(\frac {2}{5})}$

Leave your answer in terms of $\sin ^{-1}$ .