Question

Find the values of the trigonometric functions of $t$from the given information.

$\tan t=\frac{1}{4}$, the terminal point of $t$is in Quadrant III

Short answer

The values of the trigonometric functions are $\mathbf{sin}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\sqrt{\mathbf{17}}}\mathbf{,}\mathbf{cos}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\sqrt{\mathbf{17}}}\mathbf{,}\mathbf{sec}\mathit{t}\mathbf{=}\mathbf{-}\frac{\sqrt{\mathbf{17}}}{\mathbf{4}}$$\mathbf{csc}\mathit{t}\mathbf{=}\mathbf{-}\sqrt{\mathbf{17}}$, and $\mathbf{cot}\mathit{t}\mathbf{=}\mathbf{4}$.

Step-by-step solution

Step 1. State the Trigonometric identities.

The reciprocal identities are $\csc t=\frac{1}{\sin t}$, $\sec t=\frac{1}{\cos t}$, $\cot t=\frac{1}{\tan t}$,$\tan t=\frac{\sin t}{\cos t}$ and $\cot =\frac{\cos t}{\sin t}$.

The Pythagorean identities are:

1.${\sin }^{2}t+{\cos }^{2}t=1$

2. ${\tan }^{2}t+1={\sec }^{2}t$

3.${\cot }^{2}t+1={\csc }^{2}t$

Step 2. Find sect and cott. 

To find $\sec t$, substitute $\frac{1}{4}$for $\tan t$into the Pythagorean identity ${\tan }^{2}t+1={\sec }^{2}t$.

$$\begin{gathered} {\tan }^{2}t+1={\sec }^{2}t \\ {\left(\frac{1}{4}\right)}^2+1={\sec }^{2}t \\ {\sec }^{2}t=\frac{1}{16}+1 \\ \sec t=\pm \frac{\sqrt{17}}{4} \end{gathered}$$

Since the terminal point of $t$is in quadrant III, therefore, $\sec t<0$, so $\sec t=-\frac{\sqrt{17}}{4}$.

Now use reciprocal Identity $\cot t=\frac{1}{\tan t}$, substitute $\frac{1}{4}$for $\tan t$and simplify for $\cot t$.

$$\begin{gathered} \cot t=\frac{1}{\tan t} \\ =\frac{1}{\left(\frac{1}{4}\right)} \\ =4 \end{gathered}$$

Step 3. Find other trigonometric functions.

Substitute $-\frac{\sqrt{17}}{4}$for $\sec t$into reciprocal identity $\cos t=\frac{1}{\sec t}$and simplify for $\cos t$.

$\cos t=\frac{1}{\sec t}$

$$\begin{gathered} =\frac{1}{\left(-\frac{\sqrt{17}}{4}\right)} \\ =-\frac{4}{\sqrt{17}} \end{gathered}$$

Now substitute $\frac{1}{4}$for $\tan t$and $-\frac{4}{\sqrt{17}}$for $\cos t$into reciprocal identity $\tan t=\frac{\sin t}{\cos t}$and simplify for $\sin t$.

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ \frac{1}{4}=\frac{\sin t}{\left(-\frac{4}{\sqrt{17}}\right)} \\ \sin t=-\frac{4}{\sqrt{17}}\times \frac{1}{4} \\ =-\frac{1}{\sqrt{17}} \end{gathered}$$

Finally, substitute $-\frac{1}{\sqrt{17}}$for $\sin t$into reciprocal identity $\csc t=\frac{1}{\sin t}$and simplify for $\csc t$.

$$\begin{gathered} \csc t=\frac{1}{\sin t} \\ =\frac{1}{\left(-\frac{1}{\sqrt{17}}\right)} \\ =-\sqrt{17} \end{gathered}$$