Question

assignment 9.3: double - angle, half - angle formulas
score: 2/13 answered: 1/10
question 2
if \\( \sin x=\frac{8}{9} \\), \\( x \\) in quadrant i, then find the exact answers for the following (without fin
\\( \sin (2 x)= \\)
\\( \cos (2 x)= \\)
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Explanation:

Step1: Find $\cos x$

Using the identity $\sin^{2}x+\cos^{2}x = 1$.
Given $\sin x=\frac{8}{9}$, then $\cos^{2}x=1-\sin^{2}x=1 - (\frac{8}{9})^{2}=1-\frac{64}{81}=\frac{81 - 64}{81}=\frac{17}{81}$.
Since $x$ is in quadrant I, $\cos x=\frac{\sqrt{17}}{9}$.

Step2: Calculate $\sin(2x)$

Using the double - angle formula $\sin(2x)=2\sin x\cos x$.
Substitute $\sin x=\frac{8}{9}$ and $\cos x=\frac{\sqrt{17}}{9}$ into the formula:
$\sin(2x)=2\times\frac{8}{9}\times\frac{\sqrt{17}}{9}=\frac{16\sqrt{17}}{81}$.

Step3: Calculate $\cos(2x)$

Using the double - angle formula $\cos(2x)=1 - 2\sin^{2}x$.
Substitute $\sin x=\frac{8}{9}$ into the formula:
$\cos(2x)=1-2\times(\frac{8}{9})^{2}=1 - 2\times\frac{64}{81}=1-\frac{128}{81}=\frac{81-128}{81}=-\frac{47}{81}$.

Answer:

$\sin(2x)=\frac{16\sqrt{17}}{81}$
$\cos(2x)=-\frac{47}{81}$