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

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

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

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})