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Example 1
What is the exact value of $\cos (2x)$ if $\sin x = \dfrac{1}{3}$ and $x$ is in quadrant $I$?
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$\dfrac{4\sqrt 2}{3}$
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$\dfrac{7}{9}$
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$\dfrac{2}{3}$
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$\dfrac{2\sqrt 2}{3}$
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$\dfrac{8}{9}$
Solution
- We first use the identity
$\cos(2x)=\cos^2 x - \sin^2 x$
- We now use the identity $\cos^2 x = 1 - \sin ^2 x$ to rewrite $\cos (2x)$ as follows
$\cos(2x)=1 - \sin ^2 x - \sin^2 x$
- Group like terms
$\cos(2x)=1 - 2 \sin ^2 x$
- Use $\sin x = \dfrac{1}{3}$ in the expression of $\cos (2x)$ to obtain
$\cos(2x)=1 - 2(\dfrac{1}{3})^2 = 1 - 2 \dfrac{1}{9} = 1 - \dfrac{2}{9}$
$= \dfrac{9-2}{9} = \dfrac{7}{9}$
Answer B
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