The angle theta_(1) is located in Quadrant III, and sin(theta_(1))=-(12)/(13). What is the value of cos(theta_(1)) ? Express your answer exactly. cos(theta_(1))=

Step 1: Find sin(theta_(1)): We know that sin(theta_(1)) = -(12/13). Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find the value of cos(theta_(1)). Substitute -(12/13) for sin(theta_(1)) in sin²(theta_(1)) + cos²(theta_(1)) = 1. (-12/13)² + cos²(theta_(1)) = 1.Step 2: Use Pythagorean identity: Simplify (-12/13)² + cos²(theta_(1)) = 1 to find the value of cos²(theta_(1)). (144/169) + cos²(theta_(1)) = 1 cos²(theta_(1)) = 1 - 144/169 cos²(theta_(1)) = 169/169 - 144/169 cos²(theta_(1)) = 25/169Step 3: Substitute sin(theta_(1)): Find the square root of cos²(theta_(1)) to get cos(theta_(1)). cos(theta_(1)) = ±√(25/169) cos(theta_(1)) = ±5/13Step 4: Simplify the equation: Determine the sign of cos(theta_(1)) based on the quadrant in which theta_(1) is located. Since theta_(1) is in Quadrant III, both sine and cosine are negative. Therefore, cos(theta_(1)) is negative. cos(theta_(1)) = -5/13

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Q. The angle

\theta_{1}

is located in Quadrant III, and

\sin \left(\theta_{1}\right)=-\frac{12}{13}

\newline

What is the value of

\cos \left(\theta_{1}\right)

? Express your answer exactly.

\newline

\cos \left(\theta_{1}\right)=

Step $1$ : Find $\sin(\theta_{1})$ : We know that $\sin(\theta_{1}) = -\frac{12}{13}$ . $\newline$ Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find the value of $\cos(\theta_{1})$ . $\newline$ Substitute $-\frac{12}{13}$ for $\sin(\theta_{1})$ in $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ . $\newline$ $\left(-\frac{12}{13}\right)^2 + \cos^2(\theta_{1}) = 1$ .
Step $2$ : Use Pythagorean identity: Simplify $(-\frac{12}{13})^2 + \cos^2(\theta_{1}) = 1$ to find the value of $\cos^2(\theta_{1})$ . $\newline$ $\frac{144}{169} + \cos^2(\theta_{1}) = 1$ $\newline$ $\cos^2(\theta_{1}) = 1 - \frac{144}{169}$ $\newline$ $\cos^2(\theta_{1}) = \frac{169}{169} - \frac{144}{169}$ $\newline$ $\cos^2(\theta_{1}) = \frac{25}{169}$
Step $3$ : Substitute $\sin(\theta_{1})$ : Find the square root of $\cos^2(\theta_{1})$ to get $\cos(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = \pm\sqrt{\frac{25}{169}}$ $\newline$ $\cos(\theta_{1}) = \pm\frac{5}{13}$
Step $4$ : Simplify the equation: Determine the sign of $\cos(\theta_{1})$ based on the quadrant in which $\theta_{1}$ is located. $\newline$ Since $\theta_{1}$ is in Quadrant III, both sine and cosine are negative. $\newline$ Therefore, $\cos(\theta_{1})$ is negative. $\newline$ $\cos(\theta_{1}) = -\frac{5}{13}$