use identities to find values of the sine and cosine functions for the angle measure.
θ, given that ( cos 2 heta = \frac { - 12 } { 13 } ) and ( 0 ^ { circ }

Question

use identities to find values of the sine and cosine functions for the angle measure.
θ, given that ( cos 2 	heta = \frac { - 12 } { 13 } ) and ( 0 ^ { circ } < 	heta < 90 ^ { circ } )
( sin 	heta = \frac { 5 sqrt { 26 } } { 26 } )
(simplify your answer, including any radicals. use integers or fractions for any numbers in th
( cos 	heta = square )
(simplify your answer, including any radicals. use integers or fractions for any numbers in the

Accepted Answer

# Explanation: ## Step1: Use the double - angle identity for cosine We know that $\cos2\theta = 1 - 2\sin^{2}\theta$ and $\cos2\theta=\frac{- 12}{13}$. So, $\frac{-12}{13}=1 - 2\sin^{2}\theta$. $$ \begin{align*} 2\sin^{2}\theta&=1+\frac{12}{13}\ 2\sin^{2}\theta&=\frac{13 + 12}{13}\ 2\sin^{2}\theta&=\frac{25}{13}\ \sin^{2}\theta&=\frac{25}{26}\ \sin\theta&=\frac{5}{\sqrt{26}}=\frac{5\sqrt{26}}{26}\quad(0^{\circ}<\theta<90^{\circ},\text{so }\sin\theta>0) \end{align*} $$ Also, we know the double - angle identity $\cos2\theta=2\cos^{2}\theta - 1$. Substitute $\cos2\theta =-\frac{12}{13}$ into it. $$ \begin{align*} -\frac{12}{13}&=2\cos^{2}\theta-1\ 2\cos^{2}\theta&=1-\frac{12}{13}\ 2\cos^{2}\theta&=\frac{13-12}{13}\ 2\cos^{2}\theta&=\frac{1}{13}\ \cos^{2}\theta&=\frac{1}{26}\ \cos\theta&=\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26}\quad(0^{\circ}<\theta<90^{\circ},\text{so }\cos\theta>0) \end{align*} $$ # Answer: $\cos\theta=\frac{\sqrt{26}}{26}$

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Explanation:

Step1: Use the double - angle identity for cosine

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