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))= ◻

Apply Pythagorean Identity: Use the Pythagorean identity for sine and cosine. The Pythagorean identity states that sin^2(θ) + cos^2(θ) = 1. Since we know sin(theta_(1)) = -(12/13), we can substitute this into the identity to find cos(theta_(1)).Substitute sin(theta_(1)): Substitute the value of sin(theta_(1)) into the Pythagorean identity. sin^2(theta_(1)) + cos^2(theta_(1)) = 1 (-12/13)^2 + cos^2(theta_(1)) = 1 144/169 + cos^2(theta_(1)) = 1Solve for cos^2(theta_(1)): Solve for cos^2(theta_(1)). cos^2(theta_(1)) = 1 - 144/169 cos^2(theta_(1)) = 169/169 - 144/169 cos^2(theta_(1)) = (169 - 144)/169 cos^2(theta_(1)) = 25/169Find cos(theta_(1)): Take the square root of both sides to find cos(theta_(1)). Since theta_(1) is in Quadrant III, where cosine is negative, we take the negative square root. cos(theta_(1)) = -sqrt(25/169) cos(theta_(1)) = -5/13

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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))=

◻

The angle $\theta_{1}$ is located in Quadrant III, and $\sin \left(\theta_{1}\right)=-\frac{12}{13}$ . $\newline$ What is the value of $\cos \left(\theta_{1}\right)$ ? $\newline$ Express your answer exactly. $\newline$ $\cos \left(\theta_{1}\right)=$ $\newline$ $\square$

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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)

\newline

Express your answer exactly.

\newline

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

\newline

\square

Apply Pythagorean Identity: Use the Pythagorean identity for sine and cosine. $\newline$ The Pythagorean identity states that $\sin^2(\theta) + \cos^2(\theta) = 1$ . $\newline$ Since we know $\sin(\theta_{1}) = -\frac{12}{13}$ , we can substitute this into the identity to find $\cos(\theta_{1})$ .
Substitute $\sin(\theta_{1})$ : Substitute the value of $\sin(\theta_{1})$ into the Pythagorean identity. $\newline$ $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ $\newline$ $(-\frac{12}{13})^2 + \cos^2(\theta_{1}) = 1$ $\newline$ $\frac{144}{169} + \cos^2(\theta_{1}) = 1$
Solve for $\cos^2(\theta_{1})$ : Solve for $\cos^2(\theta_{1})$ .
$\cos^2(\theta_{1}) = 1 - \frac{144}{169}$
$\cos^2(\theta_{1}) = \frac{169}{169} - \frac{144}{169}$
$\cos^2(\theta_{1}) = \frac{169 - 144}{169}$
$\cos^2(\theta_{1}) = \frac{25}{169}$
Find $\cos(\theta_{1})$ : Take the square root of both sides to find $\cos(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant III, where cosine is negative, we take the negative square root. $\cos(\theta_{1}) = -\sqrt{\frac{25}{169}}$ $\cos(\theta_{1}) = -\frac{5}{13}$