Question

An angle $\mathrm{\theta }$ in the standard position for which the terminal side passes through the point $(-1,2)$, find the six trigonometric functions for $\mathrm{\theta }$.

Answer 1

Step-1: Find the radius:

The cartesian points can be written in terms of radius $\mathrm{r}$, and an angle $\mathrm{\theta }$,

Let $\mathrm{x}=\left(\mathrm{r}\right)\cos \mathrm{\theta }$ and $\mathrm{y}=\left(\mathrm{r}\right)\sin \mathrm{\theta }$.

Square both sides and add them:

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2=\left({\mathrm{r}}^2\right){\cos }^2\mathrm{\theta }+\left({\mathrm{r}}^2\right){\sin }^2\mathrm{\theta } \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }) \\ \begin{array}{rcl}& \Rightarrow & \end{array}({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1) \\ \begin{array}{rcl}& \Rightarrow & \end{array}{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2\left(1\right) \\ \begin{array}{rcl}& \Rightarrow & \end{array}{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{r}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \end{gathered}$$

For the point $(-1,2)$:

$$\begin{gathered} \mathrm{r}=\sqrt{{\left(-1\right)}^2+{\left(2\right)}^2} \\ \Rightarrow \mathrm{r}=\sqrt{1+4} \\ \therefore \mathrm{r}=\sqrt{5} \end{gathered}$$

Step-2: Find the trigonometric ratios:

Substitute the above calculation in the equations for $x$ and $y$,

Solve for $\cos \theta$:

$$\begin{gathered} -1=\sqrt{5}\cos \theta \\ \Rightarrow \cos \theta =\frac{-1}{\sqrt{5}} \end{gathered}$$

Solve for $\sin \theta$:

$$\begin{gathered} 2=\sqrt{5}\sin \theta \\ \Rightarrow \sin \theta =\frac{2}{\sqrt{5}} \end{gathered}$$

Solve for $\sec \theta$:

$$\begin{gathered} \sec \theta =\frac{1}{\cos \theta } \\ \Rightarrow \sec \theta =-\sqrt{5} \end{gathered}$$

Solve for $\mathrm{cosec}\theta$:

$$\begin{gathered} \mathrm{cosec}\theta =\frac{1}{\sin \theta } \\ \Rightarrow \mathrm{cosec}\theta =\frac{\sqrt{5}}{2} \end{gathered}$$

Solve for $\tan \theta$:

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \Rightarrow \tan \theta =\frac{{\displaystyle \frac{2}{\sqrt{5}}}}{{\displaystyle \frac{-1}{\sqrt{5}}}}\left[\because \sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{-1}{\sqrt{5}}\right] \\ \Rightarrow \tan \theta =-2 \end{gathered}$$

Solve for $\cot \theta$:

$$\begin{gathered} \cot \theta =\frac{1}{\tan \theta } \\ \Rightarrow \cot \theta =-12 \end{gathered}$$

Therefore, the six trigonometric functions are $\cos \theta =\frac{-1}{\sqrt{5}},\sin \theta =\frac{2}{\sqrt{5}},\sec \theta =-\sqrt{5},c\mathrm{osec}\theta =\frac{\sqrt{5}}{2},\tan \theta =-2,\cot \theta =-12$.