## Trigonometric Functions

It is important to know how to solve trigonometric functions since it usually appears as a question on the first math methods exam.

A sin or cos function can have either have infinite or no solutions to an equation. Therefore we will always be supplied with a domain over which the question asks us to find solutions. This will only require us to find a finite amount of solutions.

Over the domain of 2π there will be two solutions to each equation, unless it is a local max or min point. This can be seen in the graph of the sin and cos functions below. The tan function will always have one solution over its period, which is π for the basic function.

To solve a trigonometric function:

Step 1 is to identify the period length as that will tell you how many solutions you will get.

Step 2 is to solve the equation by taking the inverse of the trig function on the other side.

Step 3 is determining what the other solutions are over the domain by adding the appropriate amount of radians depending on the function.

The common convention is:

• Cos­-1 gives an answer over the range $[0,\pi]$
• Sin-1 gives an answer over the range $[-\frac{\pi}{2},\frac{\pi}{2}]$
• Tan-1 gives an answer over the range $[-\frac{\pi}{2},\frac{\pi}{2}]$

Once we have the answer within these conventional ranges we just need to add or minus a portion of the period to get the other solution within that range.

Solve the equation $sin(\frac{x}{2})=-\frac{1}{2}, x: [2\pi, 4\pi]$

Step 1 Determine the period.

$Period=\frac{2\pi}{\frac{1}{2}}=4\pi$

Step 2 Solve the equation

$\frac{x}{2}=sin^{-1}(-\frac{1}{2}) \newline \frac{x}{2}=-\frac{\pi}{6}$

We know that sin is negative in the third quadrant as well as the fourth where our answer is from. Therefore we need to add in the answer from the third quadrant as well.

$\frac{x}{2}=-\frac{\pi}{6},-\frac{5\pi}{6}$

$x=-\frac{\pi}{3},-\frac{5\pi}{3}$

Neither of our responses are in the correct domain. Therefore we need to add or minus the period until we have a result within the domain.

$x=-\frac{\pi}{3}+4\pi,-\frac{5\pi}{3}+4\pi\newline \newline x=\frac{7\pi}{3}, \frac{11\pi}{3}$

These are both within our required domain and so are our solutions.

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