Chapter 7, Problem 1RE

In Problems 1 and 2, 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 $\mathrm{\theta }$ in each figure:

Answer to Problem 1RE

Solution:

$\text{sin}\theta = \frac{4}{5}$

$\text{cos}\theta = \frac{3}{5}$

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

$\text{cosec}\theta = \frac{5}{4}$

$\text{sec}\theta = \frac{5}{3}$

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

Explanation of Solution

Given:

Formula used:

Pythagorean Theorem

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

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

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

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

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

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

Calculation:

$\begin{array}{l}{3}^2 + 4^2 = {\text{hypotenuse}}^2\\ 9 + 16 = 25\\ {\text{Hypotenuse}}^2 = 5^2\end{array}$

So removing square on both sides we will get

$\text{Hypotenuse} = 5$

Ratios:

$\begin{array}{l}\text{Opposite} \text{side} = 4\\ \text{Adjacent} \text{side} = 3\\ \text{Hypotenuse} = 5\end{array}$

$\text{sin}\theta = \frac{opposite side}{hypotenuse} = \frac{4}{5}$

$\text{cos}\theta = \frac{adjacent side}{hypotenuse} = \frac{3}{5}$

$\text{tan}\theta = \frac{\text{sin}\theta }{\text{cos}\theta } = \frac{opposite side}{adjacent side} = \frac{4}{3}$

$\text{cosec} = \frac{1}{\text{sin}\theta } = \frac{hypotenuse}{opposite side}=\frac{5}{4}$

$\text{sec}\theta = \frac{1}{\text{cos}\theta } = \frac{hypotenuse}{adjacent side} = \frac{5}{3}$

$\text{cot}\theta = \frac{\text{cos}\theta }{\text{sin}\theta } = \frac{adjacent side}{opposite side} = \frac{3}{4}$