Chapter 5.1, Problem 10E

To determine

The values of six trigonometric functions using $\tan x=\frac{\sqrt{3}}{3},\csc x=-2$ .

Answer to Problem 10E

  $\begin{array}{l}\sin x=-\frac{1}{2},\cos x=-\frac{\sqrt{3}}{2},\\ \tan x=\frac{\sqrt{3}}{3},\cot x=\sqrt{3},\\ \sec x=-\frac{2}{\sqrt{3}},\csc x=-2\end{array}$

Explanation of Solution

Given:

The values of expressions, $\tan x=\frac{\sqrt{3}}{3},\csc x=-2$

Concept Used:

• Reciprocal identities of trigonometric functions:

  $\begin{array}{l}\sin \theta =\frac{1}{\csc \theta },\csc \theta =\frac{1}{\sin \theta },\\ \cos \theta =\frac{1}{\sec \theta },\sec \theta =\frac{1}{\cos \theta },\\ \tan \theta =\frac{1}{\cot \theta },\cot \theta =\frac{1}{\tan \theta }\end{array}$

• Quotient Identities:

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

• In Quadrant I, all trigonometric functions are positive.
• In Quadrant II, only sine and cosecant are positive.
• In Quadrant III, only tangent and cotangent are positive.
• In Quadrant IV, only cosine and secant are positive.

Calculation:

In order tofind the values of the six trigonometric functions, use the reciprocal and quotient identities of trigonometric functions.

Using the reciprocal and quotient identities, it gives

  $\begin{array}{l}\sin x=\frac{1}{\csc x}=\frac{1}{-2}=-\frac{1}{2},\\ \tan x=\frac{\sin x}{\cos x}\\ \Rightarrow \cos x=\frac{\sin x}{\tan x}=\frac{-1/2}{\sqrt{3}/3}=-\frac{1}{2}\cdot \frac{3}{\sqrt{3}}=-\frac{\sqrt{3}}{2},\\ \cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}/3}=\frac{3}{\sqrt{3}}=\sqrt{3},\\ \sec x=\frac{1}{\cos x}=\frac{1}{-\sqrt{3}/2}=-\frac{2}{\sqrt{3}}\end{array}$

Thus, the values of the six trigonometric function is given by,

  $\begin{array}{l}\sin x=-\frac{1}{2},\cos x=-\frac{\sqrt{3}}{2},\\ \tan x=\frac{\sqrt{3}}{3},\cot x=\sqrt{3},\\ \sec x=-\frac{2}{\sqrt{3}},\csc x=-2\end{array}$