Question

Let sin 0= 4/9. Find the exact value of cos 0.

Answer 1

Statement Problem: Let;

$\sin \theta=\frac{4}{9}$

Find the exact value of;

$\cos \theta$

Solution:

In trigonometry, the ratio of the sine is defined as;

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

Thus,

$\text{opposite}=4,\text{ hypotenuse=9}$

By Pythagoras theorem, the square of the longest side (hypotenuse) is the sum of squares of the opposite and the adjacent sides.

$\begin{gathered} (\text{hypotenuse)}^2=(\text{opposite)}^2+(\text{adjacent)}^2 \\ (\text{adjacent)}^2=(\text{hypotenuse)}^2-(\text{opposite)}^2 \\ (\text{adjacent)}^2=9^2-4^2 \\ (\text{adjacent)}^2=81-16 \\ (\text{adjacent)}^2=65 \\ \text{adjacent}=\sqrt[]{65} \end{gathered}$

The ratio of the cosine is defined as;

$\begin{gathered} \cos \theta=\frac{adjacent}{hypotenuse} \\ \end{gathered}$

Hence,

$\cos \theta=\frac{\sqrt[]{65}}{9}$