use identities to find the values of the sine and cosine functions for the following angle measure.
θ, given that $ \cos 2 \theta = \frac { 56 } { 65 } $ and θ terminates in quadrant i
$ \sin \theta = \square $
(type an exact answer, using radicals as needed. rationalize all denominators.)
$ \cos \theta = \square $
(type an exact answer, using radicals as needed. rationalize all denominators.)

Question

Accepted Answer

# Explanation:
## Step1: Use the double - angle identity for cosine
The double - angle identity for cosine is $\cos2\theta = 1 - 2\sin^{2}\theta$.
Given $\cos2\theta=\frac{56}{65}$, we substitute it into the identity:
$\frac{56}{65}=1 - 2\sin^{2}\theta$.
First, solve for $\sin^{2}\theta$:
$$
\begin{align*}
2\sin^{2}\theta&=1-\frac{56}{65}\
2\sin^{2}\theta&=\frac{65 - 56}{65}\
2\sin^{2}\theta&=\frac{9}{65}\
\sin^{2}\theta&=\frac{9}{130}
\end{align*}
$$
Since $\theta$ terminates in quadrant I, $\sin\theta>0$. So $\sin\theta=\sqrt{\frac{9}{130}}=\frac{3}{\sqrt{130}}=\frac{3\sqrt{130}}{130}$.

## Step2: Use the double - angle identity for cosine (another form)
The double - angle identity for cosine is also $\cos2\theta=2\cos^{2}\theta - 1$.
Substitute $\cos2\theta = \frac{56}{65}$ into it:
$$
\begin{align*}
\frac{56}{65}&=2\cos^{2}\theta-1\
2\cos^{2}\theta&=1+\frac{56}{65}\
2\cos^{2}\theta&=\frac{65 + 56}{65}\
2\cos^{2}\theta&=\frac{121}{65}\
\cos^{2}\theta&=\frac{121}{130}
\end{align*}
$$
Since $\theta$ terminates in quadrant I, $\cos\theta>0$. So $\cos\theta=\sqrt{\frac{121}{130}}=\frac{11}{\sqrt{130}}=\frac{11\sqrt{130}}{130}$.

# Answer:
$\sin\theta=\frac{3\sqrt{130}}{130}$, $\cos\theta=\frac{11\sqrt{130}}{130}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Use the double - angle identity for cosine

Step2: Use the double - angle identity for cosine (another form)

Answer: