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Completion requirements
Determine the exact values of the trigonometric ratios for the angle \(135^\circ\).
Reference angle:
\(\begin{align}
\theta _R &= 180^\circ - \theta \\
\theta _R &= 180^\circ - 135^\circ \\
\theta _R &= 45^\circ \\
\end{align}\)
Use the exact values generated by the \(45^\circ - 45^\circ - 90^\circ\) special triangle. Because this angle terminates in Quadrant II, only sin \(\theta\) is positive.
\(x = -1\)
\(y = 1\)
\(r = \sqrt{2}\)
\(\begin{align}
\theta _R &= 180^\circ - \theta \\
\theta _R &= 180^\circ - 135^\circ \\
\theta _R &= 45^\circ \\
\end{align}\)
Use the exact values generated by the \(45^\circ - 45^\circ - 90^\circ\) special triangle. Because this angle terminates in Quadrant II, only sin \(\theta\) is positive.
\(x = -1\)
\(y = 1\)
\(r = \sqrt{2}\)
\(\begin{align}
\sin 135^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
&= \frac{{\sqrt 2 }}{2} \\
\end{align}\)
\(\begin{align}
\cos 135^\circ &= \frac{{ - 1}}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
&= - \frac{{\sqrt 2 }}{2} \\
\end{align}\)
\(\begin{align}
\tan 135^\circ &= \frac{1}{{-1}} \\
&= -1 \\
\end{align}\)
\sin 135^\circ &= \frac{1}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
&= \frac{{\sqrt 2 }}{2} \\
\end{align}\)
\(\begin{align}
\cos 135^\circ &= \frac{{ - 1}}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} \\
&= - \frac{{\sqrt 2 }}{2} \\
\end{align}\)
\(\begin{align}
\tan 135^\circ &= \frac{1}{{-1}} \\
&= -1 \\
\end{align}\)