given that $\sin\theta = \frac{45}{53}$ and $\cos\theta0$, determine the values of the sine and cosine functions for $2\theta$.
$\sin2\theta = \frac{2520}{2809}$ (type an integer or a simplified fraction.)
$\cos2\theta = \square$ (type an integer or a simplified fraction.)

Question

given that $\sin\theta = \frac{45}{53}$ and $\cos\theta>0$, determine the values of the sine and cosine functions for $2\theta$.
$\sin2\theta = \frac{2520}{2809}$ (type an integer or a simplified fraction.)
$\cos2\theta = \square$ (type an integer or a simplified fraction.)

Accepted Answer

# Explanation:
## Step1: Find $\cos\theta$
Use the identity $\sin^{2}\theta+\cos^{2}\theta = 1$. Given $\sin\theta=\frac{45}{53}$, then $\cos^{2}\theta=1-\sin^{2}\theta$.
$$
\begin{align*}
\cos^{2}\theta&=1 - (\frac{45}{53})^{2}\
&=\frac{53^{2}-45^{2}}{53^{2}}\
&=\frac{(53 - 45)(53 + 45)}{53^{2}}\
&=\frac{8\times98}{53^{2}}\
&=\frac{784}{2809}\
\cos\theta&=\frac{28}{53}\quad(\text{since }\cos\theta>0)
\end{align*}
$$
## Step2: Find $\cos2\theta$
Use the double - angle formula $\cos2\theta=1 - 2\sin^{2}\theta$ (or $\cos2\theta=2\cos^{2}\theta-1$).
Using $\cos2\theta=1 - 2\sin^{2}\theta$:
$$
\begin{align*}
\cos2\theta&=1-2\times(\frac{45}{53})^{2}\
&=1-\frac{2\times2025}{2809}\
&=\frac{2809-4050}{2809}\
&=\frac{2809 - 4050}{2809}\
&=\frac{28^{2}-45^{2}}{53^{2}}\
&=\frac{784 - 2025}{2809}\
&=\frac{- 1241}{2809}
\end{align*}
$$
Using $\cos2\theta=2\cos^{2}\theta-1$:
$$
\begin{align*}
\cos2\theta&=2\times(\frac{28}{53})^{2}-1\
&=\frac{2\times784}{2809}-1\
&=\frac{1568}{2809}-\frac{2809}{2809}\
&=\frac{1568 - 2809}{2809}\
&=\frac{- 1241}{2809}
\end{align*}
$$

# Answer:
$\cos2\theta=-\frac{1241}{2809}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find \(\cos\theta\)

Step2: Find \(\cos2\theta\)

Answer: