Question

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

$\sin t=-\frac{4}{5}$, terminal point of t is in Quadrant IV

Short answer

The values of the trigonometric functions are $\mathbf{cos}\mathit{t}\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\mathbf{tan}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{,}\mathbf{csc}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{4}}\mathbf{,}$$\mathbf{sec}\mathit{t}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}}$, and $\mathbf{cot}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}$.

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. Formula used. 

Now use Pythagorean Identity ${\sin }^{2}t+{\cos }^{2}t=1$, substitute $-\frac{4}{5}$for $\sin t$and simplify for $\cos t$.

$$\begin{gathered} {\sin }^{2}t+{\cos }^{2}t=1 \\ {\left(-\frac{4}{5}\right)}^2+{\cos }^{2}t=1 \\ \frac{16}{25}+{\cos }^{2}t-\frac{16}{25}=1-\frac{16}{25} \\ {\cos }^{2}t=\frac{9}{25} \\ \cos t=\pm \frac{3}{5} \end{gathered}$$

Now, t is given to be in quadrant IV, and quadrant IV, $\cos t>0$, therefore, $\cos t=\frac{3}{5}$.

Now substitute $-\frac{4}{5}$for $\sin t$for $\frac{3}{5}$and $\cos t$into reciprocal identity $\tan t=\frac{\sin t}{\cos t}$and simplify for $\tan t$.

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ =\frac{\left(-\frac{4}{5}\right)}{\frac{3}{5}} \\ =-\frac{4}{3} \end{gathered}$$

Step 3. Find other trigonometric functions.

Use the values of $\sin t,\cos t$and $\tan t$to find $\csc t,\sec t$and $\cot t$.

$\csc t=\frac{1}{\sin t}$

$$\begin{gathered} =\frac{1}{\left(-\frac{4}{5}\right)} \\ =-\frac{5}{4} \end{gathered}$$

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

$$\begin{gathered} =\frac{1}{\left(\frac{3}{5}\right)} \\ =\frac{5}{3} \end{gathered}$$

$\cot t=\frac{1}{\tan t}$

$$\begin{gathered} =\frac{1}{\left(-\frac{4}{3}\right)} \\ =-\frac{3}{4} \end{gathered}$$