Question

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

$\tan t=-\frac{12}{5},\sin t>0$

Short answer

The values of the trigonometric functions are $\mathbf{sin}\mathit{t}\mathbf{=}\frac{\mathbf{12}}{\mathbf{13}}\mathbf{,}\mathbf{cos}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{13}}\mathbf{,}\mathbf{sec}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{13}}{\mathbf{5}}$$\mathbf{csc}\mathit{t}\mathbf{=}\frac{\mathbf{13}}{\mathbf{12}}$,and $\mathbf{cot}\mathit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{12}}$.

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 $\sec t$, a substitute $-\frac{12}{5}$for $\tan t$into the Pythagorean identity ${\tan }^{2}t+1={\sec }^{2}t$.

$$\begin{gathered} {\tan }^{2}t+1={\sec }^{2}t \\ {\left(-\frac{12}{5}\right)}^2+1={\sec }^{2}t \\ {\sec }^{2}t=\frac{144}{25}+1 \\ \sec t=\pm \frac{13}{5} \end{gathered}$$

Since $\sin t>0$ and $\tan t<0$, therefore, the terminal point of $t$is in quadrant II, and in quadrant II $\sec t<0$ so, $\sec t=-\frac{13}{5}$.

Now use reciprocal Identity $\cot t=\frac{1}{\tan t}$, substitute $-\frac{12}{5}$for $\tan t$and simplify for $\cot t$.

$$\begin{gathered} \cot t=\frac{1}{\tan t} \\ =\frac{1}{\left(-\frac{12}{5}\right)} \\ =-\frac{5}{12} \end{gathered}$$

Step 3. Find other trigonometric functions.

Substitute $-\frac{13}{5}$for $\sec t$into reciprocal identity $\cos t=\frac{1}{\sec t}$and simplify for $\cos t$.

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

$$\begin{gathered} =\frac{1}{\left(-\frac{13}{5}\right)} \\ =-\frac{5}{13} \end{gathered}$$

Now substitute $-\frac{12}{5}$for $\tan t$and $-\frac{5}{13}$for $\cos t$into reciprocal identity $\tan t=\frac{\sin t}{\cos t}$and simplify for $\sin t$.

$$\begin{gathered} \tan t=\frac{\sin t}{\cos t} \\ -\frac{12}{5}=\frac{\sin t}{\left(-\frac{5}{13}\right)} \\ \sin t=-\frac{12}{5}\times \left(-\frac{5}{13}\right) \\ =\frac{12}{13} \end{gathered}$$

Finally, substitute $\frac{12}{13}$for $\sin t$into reciprocal identity $\csc t=\frac{1}{\sin t}$and simplify for $\csc t$.

$$\begin{gathered} \csc t=\frac{1}{\sin t} \\ =\frac{1}{\left(\frac{12}{13}\right)} \\ =\frac{13}{12} \end{gathered}$$