Chapter 5.1, Problem 7E

To determine

To calculate: The values of all six trigonometric functions.

Answer to Problem 7E

The required values of $\sin x$ , $\cos x$ , $\tan x$ , $\cot x$ and $\csc x$ are $\frac{\sqrt{21}}{5}$ , $-\frac{2}{5}$ , $-\frac{\sqrt{21}}{2}$ , $-\frac{2}{\sqrt{21}}$ and $\frac{5}{\sqrt{21}}$ respectively.

Explanation of Solution

Given information:

The function: $\sec x=-\frac{5}{2}$ , $\tan x<0$

Formula used:

Trigonometric identities:

  $\sin x=\frac{1}{\csc x}$

  $\sec x=\frac{1}{\cos x}$

  $\tan x=\frac{\sin x}{\cos x}$

  $\cot x=\frac{1}{\tan x}$

  ${\cos }^{2}x+{\sin }^{2}x=1$

Calculation:

The given function is $\sec x=-\frac{5}{2}$ .

So, $\begin{array}{c}\cos x=\frac{1}{\sec x}\\ =\frac{1}{-\frac{5}{2}}\\ =-\frac{2}{5}\end{array}$

It is known that ${\cos }^{2}x+{\sin }^{2}x=1$ .

Use the above formula to find $\sin x$ .

  $\begin{array}{c}{\sin }^{2}x=1-{\cos }^{2}x\\ {\sin }^{2}x=1-{\left(-\frac{2}{5}\right)}^2\\ \sin x=\sqrt{\frac{21}{25}}\\ \sin x=\frac{\sqrt{21}}{5}\end{array}$

Since, $\sin x=\frac{\sqrt{21}}{5}$ ;

So, $\begin{array}{c}\csc x=\frac{1}{\sin x}\\ =\frac{1}{\frac{\sqrt{21}}{5}}\\ =\frac{5}{\sqrt{21}}\end{array}$

To find the value of $\tan x$ , use the formula $\tan x=\frac{\sin x}{\cos x}$ .

  $\begin{array}{c}\tan x=\frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}}\\ =\frac{\sqrt{21}}{\cancel{5}}\times \left(-\frac{\cancel{5}}{2}\right)\\ =-\frac{\sqrt{21}}{2}\end{array}$

Also, $\begin{array}{c}\cot x=\frac{1}{\tan x}\\ =\frac{1}{-\frac{\sqrt{21}}{2}}\\ =-\frac{2}{\sqrt{21}}\end{array}$

Hence, the required values of $\sin x$ , $\cos x$ , $\tan x$ , $\cot x$ and $\csc x$ are $\frac{\sqrt{21}}{5}$ , $-\frac{2}{5}$ , $-\frac{\sqrt{21}}{2}$ , $-\frac{2}{\sqrt{21}}$ and $\frac{5}{\sqrt{21}}$ respectively.