Chapter 5.1, Problem 8E

To determine

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

Answer to Problem 8E

  $\sin x=-\frac{6}{7},\cos x=-\frac{\sqrt{13}}{7},\csc x=-\frac{7}{6},\sec x=-\frac{7}{\sqrt{13}},\tan x=\frac{6}{\sqrt{13}}\&\cot x=\frac{\sqrt{13}}{6}$

Explanation of Solution

Given:

  $\csc x=-\frac{7}{6},\tan x>0$

Concept used:

Reciprocal identities:

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

Pythagorean identity:

  ${\sin }^{2}u+{\cos }^{2}u=1$

Quotient identities:

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

Calculation:

Given,

  $\csc x=-\frac{7}{6},\tan x>0$

By using reciprocal identity, we have

  $\sin x=\frac{1}{\csc x}=\frac{1}{-\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}=-\frac{6}{7}$

Now by using Pythagorean identity, we have

  $\begin{array}{l}{\cos }^{2}x=1-{\sin }^{2}x\\ \Rightarrow {\cos }^{2}x=1-{\left(-\frac{6}{7}\right)}^2\\ \Rightarrow {\cos }^{2}x=1-\frac{36}{49}\\ \Rightarrow {\cos }^{2}x=\frac{13}{49}\\ \Rightarrow \cos x=\pm \frac{\sqrt{13}}{7}\end{array}$

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

  $\cos x=-\frac{\sqrt{13}}{7}$

Now by using reciprocal identity, we have

  $\sec x=\frac{1}{\cos x}=\frac{1}{-\raisebox{1ex}{$\sqrt{13}$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=-\frac{7}{\sqrt{13}}$

Now by using quotient identities, we have

  $\begin{array}{l}\tan x=\frac{\sin x}{\cos x}=\frac{-\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}{-\raisebox{1ex}{$\sqrt{13}$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=\frac{6}{\sqrt{13}}\\ \cot x=\frac{\cos x}{\sin x}=\frac{-\raisebox{1ex}{$\sqrt{13}$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}{-\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=\frac{\sqrt{13}}{6}\end{array}$