Chapter 8.1, Problem 16AYU

To determine

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

Answer to Problem 16AYU

$\text{sin}\theta =\frac{2}{\sqrt{7}}$

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

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

$\text{cosec}\theta =\frac{\sqrt{7}}{2}$

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

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

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

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

$3+4=7$

${\text{Hypotenuse}}^2={\left(\sqrt{7}\right)}^2$

So removing square on both sides we will get

$\text{Hypotenuse }=\sqrt{7}$

Ratios:

$\text{Opposite side }=2$

$\text{Adjacent side }=\sqrt{3}$

$\text{Hypotenuse }=\sqrt{7}$

$\text{Sin}\theta =\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{2}{\sqrt{7}}$

$\text{Cos}\theta =\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}$

$\text{Tan}\theta =\frac{\sin \theta }{\cos \theta }=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{2}{\sqrt{3}}$

$\text{Cosec}\theta =\frac{1}{\text{sin}\theta }=\frac{\text{hypotenuse}}{\text{opposite side}}=\frac{\sqrt{7}}{2}$

$\text{Sec}\theta =\frac{1}{\text{cos}\theta }=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{\sqrt{7}}{\sqrt{3}}=\sqrt{\frac{7}{3}}$

$\text{Cot}\theta =\frac{\cos \theta }{\sin \theta }=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{\sqrt{3}}{2}$