use identities to find values of the sine and cosine functions for the angle measure. \nθ, given that \\(…

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{12}{13}), we substitute it into the identity: (-\frac{12}{13}=1 - 2\sin^{2}\theta)

Step2: Solve for (\sin\theta)

First, rearrange the equation: (2\sin^{2}\theta=1+\frac{12}{13}) (2\sin^{2}\theta=\frac{13 + 12}{13}=\frac{25}{13}) (\sin^{2}\theta=\frac{25}{26}) Since (0^{\circ}<\theta<90^{\circ}), (\sin\theta>0). So (\sin\theta=\sqrt{\frac{25}{26}}=\frac{5}{\sqrt{26}}=\frac{5\sqrt{26}}{26})

Step3: Use the double - angle identity for cosine again

Another double - angle identity for cosine is (\cos2\theta = 2\cos^{2}\theta-1) Substitute (\cos2\theta =-\frac{12}{13}) into it: (-\frac{12}{13}=2\cos^{2}\theta - 1)

Step4: Solve for (\cos\theta)

Rearrange the equation: (2\cos^{2}\theta=1-\frac{12}{13}) (2\cos^{2}\theta=\frac{13-12}{13}=\frac{1}{13}) (\cos^{2}\theta=\frac{1}{26}) Since (0^{\circ}<\theta<90^{\circ}), (\cos\theta>0). So (\cos\theta=\sqrt{\frac{1}{26}}=\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26})

Answer:

(\sin\theta=\frac{5\sqrt{26}}{26}), (\cos\theta=\frac{\sqrt{26}}{26})