11.1 Graphs, trigonometric identities, and solving trigonometric (Page 9/12)

Page 9 / 12

Solve for $θ$ :

sin θ = 0, 3

Determine in which quadrants the solution lies
We look at the sign of the trigonometric function. $sin θ$ is given as a positive amount ( $0, 3$ ). Reference to the CAST diagram shows that sine is positive in the first and second quadrants.

S A

T C
Determine the reference angle
The small angle $θ$ is the angle returned by the calculator:

$\begin{matrix} sin θ & = & 0, 3 \\ \Rightarrow θ & = & arcsin 0, 3 \\ \Rightarrow θ & = & 17, 46^{\circ} \end{matrix}$
Determine the general solution
Our solution lies in quadrants I and II. We therefore use $θ$ and $180^{\circ} - θ$ , and add the $360^{\circ} \cdot n$ for the periodicity of sine.

$180^{\circ} - θ$ $θ$

$180^{\circ} + θ$ $360^{\circ} - θ$

$\begin{matrix} I : θ & = & 17, 46^{\circ} + 360^{\circ} \cdot n, n \in Z \\ II : θ & = & 180^{\circ} - 17, 46^{\circ} + 360^{\circ} \cdot n, n \in Z \\ = & 162, 54^{\circ} + 360^{\circ} \cdot n, n \in Z \end{matrix}$

This is called the general solution .
Find the specific solutions
We can then find all the values of $θ$ by substituting $n = ..., - 1, 0, 1, 2, ...$ etc. For example,If $n = 0, θ = 17, 46^{\circ}; 162, 54^{\circ}$ If $n = 1, θ = 377, 46^{\circ}; 522, 54^{\circ}$ If $n = - 1, θ = - 342, 54^{\circ}; - 197, 46^{\circ}$ We can find as many as we like or find specific solutions in a given interval by choosing more values for $n$ .

General solution using periodicity

Up until now we have only solved trigonometric equations where the argument (the bit after the function, e.g. the $θ$ in $cos θ$ or the $(2 x - 7)$ in $tan (2 x - 7)$ ), has been $θ$ . If there is anything more complicated than this we need to be a little more careful. Let us try to solve $tan (2 x - 10^{\circ}) = 2, 5$ in the range $- 360^{\circ} \leq x \leq 360^{\circ}$ . We want solutions for positive tangent so using our CAST diagram we know to look in the 1 $^{st}$ and 3 $^{rd}$ quadrants. Our calculator tells us that $arctan (2, 5) = 68, 2^{\circ}$ . This is our reference angle. So to find the general solution we proceed as follows:

\begin{matrix} tan (2 x - 10^{\circ}) & = & 2, 5 [68, 2^{\circ}] \\ I : 2 x - 10^{\circ} & = & 68, 2^{\circ} + 180^{\circ} \cdot n \\ 2 x & = & 78, 2^{\circ} + 180^{\circ} \cdot n \\ x & = & 39, 1^{\circ} + 90^{\circ} \cdot n, n \in Z \end{matrix}

This is the general solution. Notice that we added the $10^{\circ}$ and divided by 2 only at the end. Notice that we added $180^{\circ} \cdot n$ because the tangent has a period of $180^{\circ}$ . This is also divided by 2 in the last step to keep the equation balanced. We chose quadrants I and III because $tan$ was positive and we used the formulae $θ$ in quadrant I and $(180^{\circ} + θ)$ in quadrant III. To find solutions where $- 360^{\circ} < x < 360^{\circ}$ we substitue integers for $n$ :

$n = 0$ ; $x = 39, 1^{\circ}$ ; $129, 1^{\circ}$
$n = 1$ ; $x = 129, 1^{\circ}$ ; $219, 1^{\circ}$
$n = 2$ ; $x = 219, 1^{\circ}$ ; $309, 1^{\circ}$
$n = 3$ ; $x = 309, 1^{\circ}$ ; $399, 1^{\circ}$ (too big!)
$n = - 1$ ; $x = - 50, 9^{\circ}$ ; $39, 1^{\circ}$
$n = - 2$ ; $x = - 140, 1^{\circ}$ ; $- 50, 9^{\circ}$
$n = - 3$ ; $x = - 230, 9^{\circ}$ ; $- 140, 9^{\circ}$
$n = - 4$ ; $x = - 320, 9^{\circ}$ ; $- 230, 9^{\circ}$

Solution: $x = - 320, 9^{\circ}; - 230^{\circ}; - 140, 9^{\circ}; - 50, 9^{\circ}; 39, 1^{\circ}; 129, 1^{\circ}; 219, 1^{\circ}$ and $309, 1^{\circ}$

Linear trigonometric equations

Just like with regular equations without trigonometric functions, solving trigonometric equations can become a lot more complicated. You should solve these just like normal equations to isolate a single trigonometric ratio. Then you follow the strategy outlined in the previous section.

Write down the general solution for $3 cos (θ - 15^{\circ}) - 1 = - 2, 583$

Use what we have learnt so far:
$\begin{matrix} 3 cos (θ - 15^{\circ}) - 1 & = & - 2, 583 \\ 3 cos (θ - 15^{\circ}) & = & - 1, 583 \\ cos (θ - 15^{\circ}) & = & - 0, 5276 . . . \\ reference angle : (θ - 15^{\circ}) & = & 58, 2^{\circ} \\ II : θ - 15^{\circ} & = & 180^{\circ} - 58, 2^{\circ} + 360^{\circ} \cdot n, n \in Z \\ θ & = & 136, 8^{\circ} + 360^{\circ} \cdot n, n \in Z \\ III : θ - 15^{\circ} & = & 180^{\circ} + 58, 2^{\circ} + 360^{\circ} \cdot n, n \in Z \\ θ & = & 253, 2^{\circ} + 360^{\circ} \cdot n, n \in Z \end{matrix}$

Quadratic and higher order trigonometric equations

The simplest quadratic trigonometric equation is of the form

{sin}^{2} x - 2 = - 1, 5

This type of equation can be easily solved by rearranging to get a more familiar linear equation

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Source: OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1

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