Question

use identities to find values of the sine and cosine functions for the angle measure.
θ, given that ( cos 2 \theta=\frac{56}{65} ) and ( 90^{circ}<\theta<180^{circ} )
( sin \theta=\frac{3 sqrt{130}}{130} )
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)
( cos \theta=square )
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Use the double - angle identity for cosine

The double - angle identity for cosine is \(\cos2\theta = 1 - 2\sin^{2}\theta=2\cos^{2}\theta - 1\). We can also use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\). Since we know \(\cos2\theta=\frac{56}{65}\), and we want to find \(\cos\theta\), we use the identity \(\cos2\theta=2\cos^{2}\theta - 1\).
Rearrange the identity to solve for \(\cos^{2}\theta\):
\(\cos^{2}\theta=\frac{1 + \cos2\theta}{2}\)
Substitute \(\cos2\theta=\frac{56}{65}\) into the formula:
\(\cos^{2}\theta=\frac{1+\frac{56}{65}}{2}=\frac{\frac{65 + 56}{65}}{2}=\frac{121}{130}\)

Step2: Determine the sign of \(\cos\theta\)

Given that \(90^{\circ}<\theta<180^{\circ}\), \(\theta\) is in the second quadrant. In the second quadrant, the cosine function is negative.
Take the square root of \(\cos^{2}\theta=\frac{121}{130}\). So \(\cos\theta=-\sqrt{\frac{121}{130}}\)
Rationalize the denominator: \(\cos\theta=-\frac{11}{\sqrt{130}}=-\frac{11\sqrt{130}}{130}\)

Answer:

\(\cos\theta =-\frac{11\sqrt{130}}{130}\)