Chapter 5.1, Problem 12E

To determine

To find the values of all six trigonometric functions by using given conditions.

Answer to Problem 12E

  $\sin x=-\frac{4}{\sqrt{65}},\cos x=-\frac{7}{\sqrt{65}},\csc x=-\frac{\sqrt{65}}{4},\sec x=-\frac{\sqrt{65}}{7},\tan x=\frac{4}{7}\&\cot x=\frac{7}{4}$

Explanation of Solution

Given:

  $\cot x=\frac{7}{4},\sin x<0$

Concept used:

Reciprocal identities:

  $\tan u=\frac{1}{\cot u},\csc u=\frac{1}{\sin u},\cos u=\frac{1}{\sec u}$

Pythagorean identity:

  $\begin{array}{l}{\sin }^{2}u+{\cos }^{2}u=1\\ {\sec }^{2}u=1+{\tan }^{2}u\end{array}$

Calculation:

Given, $\cot x=\frac{7}{4},\sin x<0$

By using reciprocal identity, we have

  $\tan x=\frac{1}{\cot x}=\frac{1}{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{4}{7}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\sec }^{2}x=1+{\tan }^{2}x\\ \Rightarrow {\sec }^{2}x=1+{\left(\frac{4}{7}\right)}^2\\ \Rightarrow {\sec }^{2}x=1+\frac{16}{49}\\ \Rightarrow {\sec }^{2}x=\frac{65}{49}\\ \Rightarrow \sec x=\pm \frac{\sqrt{65}}{7}\end{array}$

Since $\cot x>0\&\sin x<0$ , it follows that $x$ lies in quadrant III. Moreover, $\sec x$ is negative in quadrant III. Therefore,

  $\sec x=-\frac{\sqrt{65}}{7}$

Now by using reciprocal identity, we have

  $\cos x=\frac{1}{\sec x}=\frac{1}{-\raisebox{1ex}{$\sqrt{65}$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=-\frac{7}{\sqrt{65}}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\sin }^{2}x=1-{\cos }^{2}x\\ \Rightarrow {\sin }^{2}x=1-{\left(-\frac{7}{\sqrt{65}}\right)}^2\\ \Rightarrow {\sin }^{2}x=1-\frac{49}{65}\\ \Rightarrow {\sin }^{2}x=\frac{16}{65}\\ \Rightarrow \sin x=\pm \frac{4}{\sqrt{65}}\end{array}$

Since $\cot x>0\&\sin x<0$ , it follows that $x$ lies in quadrant III. Moreover, $\sin x$ is negative in quadrant III. Therefore,

  $\sin x=-\frac{4}{\sqrt{65}}$

Now by using reciprocal identity, we have

  $\csc x=\frac{1}{\sin x}=\frac{1}{-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$\sqrt{65}$}\right.}=-\frac{\sqrt{65}}{4}$