Question

practice assignment 9.3 double - angle and half - angle identities
score: 0/6 answered: 0/6
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question 1
if $sin x=\frac{3}{4}$, $x$ in quadrant i, then find (without finding $x$):
$sin(2x)=$
$cos(2x)=$
$\tan(2x)=$
question help: video 1 video 2

Explanation:

Step1: Find $\cos x$

Using the identity $\sin^{2}x+\cos^{2}x = 1$.
Given $\sin x=\frac{3}{4}$, then $\cos^{2}x=1-\sin^{2}x=1 - (\frac{3}{4})^{2}=1-\frac{9}{16}=\frac{7}{16}$.
Since $x$ is in quadrant I, $\cos x=\frac{\sqrt{7}}{4}$.

Step2: Find $\sin(2x)$

Using the double - angle formula $\sin(2x)=2\sin x\cos x$.
Substitute $\sin x = \frac{3}{4}$ and $\cos x=\frac{\sqrt{7}}{4}$ into the formula:
$\sin(2x)=2\times\frac{3}{4}\times\frac{\sqrt{7}}{4}=\frac{3\sqrt{7}}{8}$.

Step3: Find $\cos(2x)$

Using the double - angle formula $\cos(2x)=1 - 2\sin^{2}x$.
Substitute $\sin x=\frac{3}{4}$ into the formula:
$\cos(2x)=1-2\times(\frac{3}{4})^{2}=1 - 2\times\frac{9}{16}=1-\frac{9}{8}=-\frac{1}{8}$.

Step4: Find $\tan(2x)$

Using the formula $\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$.
Substitute $\sin(2x)=\frac{3\sqrt{7}}{8}$ and $\cos(2x)=-\frac{1}{8}$ into the formula:
$\tan(2x)=\frac{\frac{3\sqrt{7}}{8}}{-\frac{1}{8}}=- 3\sqrt{7}$.

Answer:

$\sin(2x)=\frac{3\sqrt{7}}{8}$; $\cos(2x)=-\frac{1}{8}$; $\tan(2x)=-3\sqrt{7}$