use identities to find the values of the sine and cosine functions for the following angle measure.\nθ…

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})