Example 5
Completion requirements
| Example 5 |
Multimedia |
Given sin \(\theta = \frac{5}{6}\) and \(\theta\) is an angle in standard position that terminates in Quadrant II, determine the exact values of cos \(\theta\) and tan \(\theta\).
Step 1: Sketch the angle and reference triangle.
Step 2: Determine the value of \(x\).
\(\begin{align}
r^2 &= x^2 + y^2 \\
\left( 6 \right)^2 &= x^2 + \left( 5 \right)^2 \\
36 &= x^2 + 25 \\
11 &= x^2 \\
\pm \sqrt {11} &= x \\
\end{align}\)
Because the terminal arm is in Quadrant II, the \(x\)-value is negative; therefore, \(x = -\sqrt{11}\).
Step 3: Determine the exact values of cos \(\theta\) and tan \(\theta\).
\(\begin{align}
x &= -\sqrt{11} \\
y &= 5 \\
r &= 6 \\
\end{align}\)
\(\begin{align}
\cos \theta &= \frac{x}{r} \\
&= \frac{{ - \sqrt {11} }}{6} \\
&= - \frac{{\sqrt {11} }}{6} \\
\tan \theta &= \frac{y}{x} \\
&= \frac{5}{{ - \sqrt {11} }} \cdot \frac{{\sqrt {11} }}{{\sqrt {11} }} \\
&= - \frac{{5\sqrt {11} }}{{11}} \\
\end{align}\)
Step 2: Determine the value of \(x\).
\(\begin{align}
r^2 &= x^2 + y^2 \\
\left( 6 \right)^2 &= x^2 + \left( 5 \right)^2 \\
36 &= x^2 + 25 \\
11 &= x^2 \\
\pm \sqrt {11} &= x \\
\end{align}\)
Because the terminal arm is in Quadrant II, the \(x\)-value is negative; therefore, \(x = -\sqrt{11}\).
Step 3: Determine the exact values of cos \(\theta\) and tan \(\theta\).
\(\begin{align}
x &= -\sqrt{11} \\
y &= 5 \\
r &= 6 \\
\end{align}\)
\(\begin{align}
\cos \theta &= \frac{x}{r} \\
&= \frac{{ - \sqrt {11} }}{6} \\
&= - \frac{{\sqrt {11} }}{6} \\
\tan \theta &= \frac{y}{x} \\
&= \frac{5}{{ - \sqrt {11} }} \cdot \frac{{\sqrt {11} }}{{\sqrt {11} }} \\
&= - \frac{{5\sqrt {11} }}{{11}} \\
\end{align}\)