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\noindent
{\large\bf Inverse Trig Functions:}

None of the trig functions are one-to-one, so none
of them have inverses. However, we need inverses to 
solve equations! How do we solve this problem?
The answer is that we restrict the domains of
the trig functions in such a way as to obtain 
one-to-one functions. Then we only discuss
inverses of functions with restricted domain.
See the pictures.

\bigskip

\noindent
{\bf Defining Properties:}

As you know, the defining property for in inverse function
is that $f^{-1}\circ f(x) = x$ and 
$f\circ f^{-1}(x) = x$ for all $x$. Since we have restricted
our domains, we can only insist on
these properties for certain values of $x$.
(Notation: the inverse of a trig function such as
$\sin(x)$ is written $\arcsin(x)$ or $\sin^{-1}(x)$.)


\medskip

\noindent
$\arcsin(\sin(x)) = x$ for $x\in [-\pi/2,\pi/2]$ and 
$\sin(\arcsin(x)) = x$ for $x\in [-1,1]$ and 

\medskip

\noindent
$\arccos(\cos(x)) = x$ for $x\in [0,\pi]$ and 
$\cos(\arccos(x)) = x$ for $x\in [-1,1]$ and 

\medskip

\noindent
$\arctan(\tan(x)) = x$ for $x\in [-\pi/2,\pi/2]$ and 
$\tan(\arctan(x)) = x$ for $x\in [-\infty,\infty]$ and 

\bigskip

\noindent
{\bf Exercises.}

\begin{enumerate}
\item[(1)] Why is $\displaystyle
\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right)\neq
\frac{5\pi}{6}$?
What is $\displaystyle
\arcsin\left(\sin\left(\frac{5\pi}{6}\right)\right)$?
\item[(2)] Find $\displaystyle
\sin\left(\arccos\left(\frac{1}{2}\right)\right)$.
\item[(3)] Show that $\displaystyle
\sin\left(2\arcsin\left(x\right)\right) = 2x\sqrt{1-x^2}$ when 
$x\in[-1,1]$.
\item[(4)] True or False: $\displaystyle
\tan^{-1}(x) = \frac{\sin^{-1}(x)}{\cos^{-1}(x)}$?
What about 
\[\tan^{-1}(x) = \left(\frac{\sin(x)}{\cos(x)}\right)^{-1}
= \left(\frac{\cos(x)}{\sin(x)}\right) = \cot(x)?\]
\end{enumerate}

\bigskip


\noindent
{\large\bf Trigonometric Equations:}

Calculations with trig functions 
often lead to equations that we have to solve.
Sometimes they are simple equations like
$\sin(x) = \frac{1}{2}$, or they may be more
complicated equations like
$4\cos(2x) + 4\sin^4(x) = 1$.
Solving equations requires combining all your
knowledge of trig functions: definitions,
properties of trig graphs, special values of trig
functions, identities satisfied by trig functions,
and values of inverse trig functions.
The best way to learn to solve trig equations
is by practicing, so let's start!

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\noindent
{\bf Exercises.}
Find all intersection points for the 
following pairs of graphs.

\begin{enumerate}
\item[(5)] $y=\sin(x)$ and $y=\cos(x)$.
\item[(6)] $y=\sin(x)$ and $y=\tan(x)$.
\item[(7)] $y=\tan(x)$ and $y=\cot(2x)$.
\item[(8)] $y=\sin(x)$ and $y=\sin(2x)$.
\end{enumerate}


\bigskip


\noindent
{\bf More Exercises.}
Solve the following trig equations.

\begin{enumerate}
\item[(9)] $1+\sin(x)=\cos(x)$
\item[(10)] $\sin(x) + \cos(x) = 0$.
\item[(11)] $4\sin^3(x) - 2\sin^2(x) - 2\sin(x) + 1 = 0$.
(Hint: Try to solve $x^3-2x^2-2x+1=0$ first.)
\item[(12)] $\tan(3x) + 1 = \sec(3x)$. (Find only
the solutions
in the interval $[0,2\pi)$.)
\end{enumerate}









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