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

Let sin 0= 4/9. Find the exact value of cos 0.-example-1

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Statement Problem: Let;


\sin \theta=(4)/(9)

Find the exact value of;


\cos \theta

Solution:

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


\sin \theta=(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=(adjacent)/(hypotenuse) \\ \end{gathered}

Hence,


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

User Ricky Sahu
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