Given any point \(P\,(x,y)\) on the terminal arm of a standard position angle, we can determine the value of any or all of the six trigonometric ratios.
If that point is also on the unit circle, the process of determining the value of any or all of the six trigonometric ratios is easier.
Point \(P\, \left( -\dfrac{3}{5} , -\dfrac{4}{5} \right)\) is on the terminal arm of standard position angle \(\theta\). Determine the exact values of each of the six trigonometric ratios.
Solution
A sketch is generally helpful.
First, we calculate the value of \(r\).
\(r^2 = x^2 + y^2 = \left( -\dfrac{3}{5} \right)^2 + \left( -\dfrac{4}{5} \right)^2 = \dfrac{9}{25} + \dfrac{16}{25} = 1\)
and \(r = 1\) since \(r \gt 0\). Since \(r = 1\), the point \(P\,\left( -\dfrac{3}{5}, -\dfrac{4}{5} \right)\) is on the unit circle and \((x,y) = (\cos(\theta), \sin(\theta))\). Then,
| \(\sin(\theta) = -\dfrac{4}{5}\) |
\(\cos(\theta) = -\dfrac{3}{5}\) |
\(\tan(\theta) = \dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{4}{3}\) |
| \(\csc(\theta) = \dfrac{1}{\sin(\theta)} = -\dfrac{5}{4}\) |
\(\sec(\theta) = \dfrac{1}{\cos(\theta)} = -\dfrac{5}{3}\) |
\(\cot(\theta) = \dfrac{1}{\tan(\theta)} = \dfrac{3}{4}\) |