Chapter 8.1, Problem 15AYU

In Problems 9-18, find the exact value of the six trigonometric functions of the angle $\theta$ in each figure.

To determine

To find: The exact value of the six trigonometric functions of the angle $\theta$ in the figure:

Answer to Problem 15AYU

$\text{sin}\theta =\sqrt{\frac{2}{3}}$

$\text{cos}\theta =\frac{1}{\sqrt{3}}$

$\text{tan}\theta =\sqrt{2}$

$\text{cosec}\theta =\frac{4}{3}$

$\text{sec}\theta =\frac{4}{\sqrt{7}}$

$\text{cot}\theta =\frac{\sqrt{7}}{3}$

Explanation of Solution

Given:

Formula used:

Pythagorean Theorem

$\text{Sin}\theta =\frac{\text{opposite side}}{\text{hypotenuse}}$

$\text{Cos}\theta =\frac{\text{adjacent side}}{\text{hypotenuse}}$

$\text{Tan}\theta =\frac{\sin \theta }{\cos \theta }=\frac{\text{opposite side}}{\text{adjacent side}}$

$\text{Cosec}\theta =\frac{1}{\text{sin}\theta }=\frac{\text{hypotenuse}}{\text{opposite side}}$

$\text{Sec}\theta =\frac{1}{\text{cos}\theta }=\frac{\text{hypotenuse}}{\text{adjacent side}}$

$\text{Cot}\theta =\frac{\cos \theta }{\sin \theta }=\frac{\text{adjacent side}}{\text{opposite side}}$

Calculation:

From the above fig. by applying Pythagorean Theorem

${\left(\sqrt{2}\right)}^2+1^2={\text{ hypotenuse}}^2$

$2+1=3$

${\text{Hypotenuse}}^2={\left(\sqrt{3}\right)}^2$

So removing square on both sides we will get

$\text{Hypotenuse }=\sqrt{3}$

Ratios:

$\text{Opposite side }=\sqrt{2}$

$\text{Adjacent side }=1$

$\text{Hypotenuse }=\sqrt{3}$

$\text{Sin}\theta =\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}$

$\text{Cos}\theta =\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{1}{\sqrt{3}}$

$\text{Tan}\theta =\frac{\sin \theta }{\cos \theta }=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{\sqrt{2}}{1}=\sqrt{2}$

$\text{Cosec}\theta =\frac{1}{\text{sin}\theta }=\frac{\text{hypotenuse}}{\text{opposite side}}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}$

$\text{Sec}\theta =\frac{1}{\text{cos}\theta }=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{\sqrt{3}}{1}=\sqrt{3}$

$\text{Cot}\theta =\frac{\cos \theta }{\sin \theta }=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{1}{\sqrt{2}}$