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Precalculus Review Materials

Jane Butterfield, Gary MacGillivray

Section 5.4 Finding the Values of Trigonometric Functions

Give the values for one trig function, we can find the values for the other five trig functions. There are two ways to do this.
Method 1: Use identities.
Given \(\theta\) in Quad III and \(\cot\theta=2\text{,}\) find the values of the remaining trig functions.
Solution.
First, cotangent and tangent are reciprocals, so
\begin{equation*} \tan \theta = \frac{1}{\cot\theta}=\frac{1}{2}. \end{equation*}
Next use the identity \(\cot^2\theta+1\csc^2\theta\) to get \(\csc\theta\text{:}\)
\begin{align*} \cot^2\theta + 1 = \amp \csc^2\theta \\ 2^2 + 1 = \amp \csc^2\theta\\ \csc\theta = \amp \pm\sqrt{5}\\ \csc\theta = \amp \sqrt{5} \end{align*}
since \(\theta\) is in Quad III.
Cosecant and sine are reciprocals, so
\begin{equation*} \sin\theta = \frac{1}{\csc\theta} = - \frac{1}{\sqrt{5}}. \end{equation*}
Using \(\sin^2\theta+\cos^2\theta=1\text{,}\) we get
\begin{align*} \cos^2\theta + \left(-\frac{1}{\sqrt{5}}\right)^2 = \amp 1\\ \cos^2\theta = \amp \frac{4}{5}\\ \cos \theta = \amp \pm \frac{2}{\sqrt{5}} \\ \cos \theta = \amp -\frac{2}{\sqrt{5}} \end{align*}
since \(\theta\) is in Quad III.
Cosine and secant are reciprocals, so
\begin{equation*} \sec\theta = \frac{1}{\cos\theta} = - \frac{\sqrt{5}}{2}. \end{equation*}
This gives us all six trig function values.
Method 2: Use a right triangle with \(0 \text{ radians} \lt \theta \lt \frac{\pi}{2}\text{ radians}\text{.}\)
A right triangle with one acute angle labled \(\theta\text{,}\) the leg adjanct with that angle labled “adjacent”, the other leg labled “opposite”, and the hypotenuse labled “hypotenuse”.
Figure 5.4.2.
This gives
\begin{align*} \amp \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \amp \amp \cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}} \amp \amp \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \end{align*}
One can remember these equations using the acronym “SOH CAH TOA”, where S is sine, O is opposite, H is hypotenuse, C is cosine, A is adjacent, and T is tangent.
(In the previous triangle, we assumed \(\theta\) was in Quad I. However, by using symmetry, we can assume \(\theta\) is in any quadrant. But be careful of the signs of the trig functions when \(\theta\) is in Quad II, III, or IV!)
We will now work the previous example again, using Method 2.
Given \(\theta\) in Quad III and \(\cot\theta=2\text{,}\) find the values of the remaining trig functions.
Solution.
We then have the following triangle.
A right triangle with leg adjacent to \(\theta\) having length 2 and the one opposite having length 1. Calculations pictured to the right: \(1^2+2^2=h^2\text{,}\) so \(h=\sqrt{5}\text{.}\)
Figure 5.4.4.
Next note that \(\theta\) is in Quad III. Thus
\begin{align*} \cos \theta \amp = - \frac{\text{adj}}{\text{hyp}} = -\frac{2}{\sqrt{5}} \amp \sec \theta \amp = - \frac{1}{\cos\theta} = -\frac{\sqrt{5}}{2}\\ \sin \theta \amp = - \frac{\text{opp}}{\text{hyp}} = -\frac{21}{\sqrt{5}} \amp \csc \theta \amp = - \frac{1}{\sin\theta} = -\sqrt{5}\\ \tan \theta \amp = \frac{\text{opp}}{\text{adj}} = \frac{1}{2} \end{align*}
This gives us the values of all six trig functions.
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