Chapter 5.1, Problem 17E

To determine

The values of six trigonometric functions using $\tan \theta =2,\sin \theta <0$ .

Answer to Problem 17E

  $\begin{array}{l}\sin \theta =-\frac{2\sqrt{5}}{5},\cos \theta =-\frac{\sqrt{5}}{5},\\ \tan \theta =2,\cot \theta =\frac{1}{2},\\ \sec \theta =-\sqrt{5},\csc \theta =-\frac{\sqrt{5}}{2}\end{array}$

Explanation of Solution

Given:

The values of expressions, $\tan \theta =2,\sin \theta <0$

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.

Since tan is positive and sine is negative, so the angle should be in Quadrant III.

From the trigonometric ratios, we know that $\tan =\frac{\text{Opposite}}{\text{Adjacent}}=\frac{2}{1}$ .

So, using the Pythagorean theorem, it gives

  $\begin{array}{c}\text{Hypotenus}=\sqrt{{\left(\text{adjacent}\right)}^2+{\left(\text{opposite}\right)}^2}\\ =\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

Since the angle is in third quadrant, so only tangent and cotangent will be positive, rest other trigonometric function will be negative.

So, using the reciprocal and quotient identities, it gives

  $\begin{array}{l}\sin \theta =\frac{\text{Opposite}}{\text{Hypotenus}}=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5},\\ \cos \theta =\frac{\text{Adjacent}}{\text{Hypotenus}}=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5},\\ \cot \theta =\frac{1}{\tan \theta }=\frac{1}{2},\\ \sec \theta =\frac{1}{\cos \theta }=\frac{1}{-1/\sqrt{5}}=-\sqrt{5},\\ \csc \theta =\frac{1}{\sin \theta }=\frac{1}{-2/\sqrt{5}}=-\frac{\sqrt{5}}{2}\end{array}$

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

  $\begin{array}{l}\sin \theta =-\frac{2\sqrt{5}}{5},\cos \theta =-\frac{\sqrt{5}}{5},\\ \tan \theta =2,\cot \theta =\frac{1}{2},\\ \sec \theta =-\sqrt{5},\csc \theta =-\frac{\sqrt{5}}{2}\end{array}$