Applications of the Unit Circle

The identify $\cos^2 \theta + \sin^2 \theta = 1$ is required for finding trigonometric ratios.

Example 1

Find exactly the possible values of $\cos \theta$ for $\sin \theta = \dfrac{5}{8}$.

\( \begin{align} \displaystyle
\cos^2 \theta + \sin^2 \theta &= 1 \\
\cos^2 \theta + \sin^2 \dfrac{5}{8} &= 1 \\
\cos^2 \theta + \dfrac{25}{64} &= 1 \\
\cos^2 \theta &= \dfrac{39}{64} \\
\therefore \cos \theta &= \pm \dfrac{\sqrt{39}}{8}
\end{align} \)

Example 2

If $\sin \theta = -\dfrac{2}{3}$ and $\pi \lt \theta \lt \dfrac{3 \pi}{2}$, find the exact values of $\cos \theta$ and $\tan \theta$.

The angle $\theta$ is in quadrant 3, so $\cos \theta \lt 0$ and $\tan \theta \gt 0$.
\( \require{AMSsymbols} \begin{align} \displaystyle \require{color}
\cos^2 \theta + \sin^2 \theta &= 1 \\
\cos^2 \theta + \dfrac{4}{9} &= 1 \\
\cos^2 \theta &= \dfrac{5}{9} \\
\cos \theta &= \pm \dfrac{\sqrt{5}}{3} \\
\therefore \cos \theta &= -\dfrac{\sqrt{5}}{3} &\color{red} \text{angle in quadrant 3}\\
\tan \theta &= \dfrac{\sin \theta}{\cos \theta} \\
&= \dfrac{-\dfrac{2}{3}}{-\dfrac{\sqrt{5}}{3}} \\
&= \dfrac{2}{\sqrt{5}} \\
\therefore \cos \theta &= -\dfrac{\sqrt{5}}{3}, \tan \theta = \dfrac{2}{\sqrt{5}}
\end{align} \)

Example 3

If $\tan \theta = -3$ and $\dfrac{3 \pi}{2} \lt \theta \lt 2 \pi$, find the exact values of $\sin \theta$ and $\cos \theta$.

The angle $\theta$ is in quadrant 4, so $\sin \theta \gt 0$ and $\cos \theta \lt 0$.
\( \require{AMSsymbols} \begin{align} \displaystyle \require{color}
\tan \theta &= \dfrac{\sin \theta}{\cos \theta} = -3 \\
\sin \theta &= -3 \cos \theta \\
\cos^2 \theta + \sin^2 \theta &= 1 \\
\cos^2 \theta + (-3\cos \theta)^2 &= 1 \\
\cos^2 \theta + 9\cos^2 \theta &= 1 \\
10 \cos^2 \theta &= 1 \\
\cos \theta &= \pm \dfrac{1}{\sqrt{10}} \\
\cos \theta &= \dfrac{1}{\sqrt{10}} &\color{red} \text{angle in quadrant 4}\\
\sin \theta &= -3 \cos \theta \\
&= -3 \times \dfrac{1}{\sqrt{10}} \\
&= \dfrac{-3}{\sqrt{10}} \\
\therefore \sin \theta &= \dfrac{-3}{\sqrt{10}}, \cos \theta = \dfrac{1}{\sqrt{10}}
\end{align} \)

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