The angle theta_(1) is located in Quadrant II, and cos(theta_(1))=-(12)/(19). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Pythagorean identity: We know that for any angle θ, sin²(θ) + cos²(θ) = 1 (Pythagorean identity). Given that cos(theta_(1)) = -(12/19), we can find sin(theta_(1)) by rearranging the Pythagorean identity to solve for sin²(theta_(1)). sin²(theta_(1)) = 1 - cos²(theta_(1))Rearranging the Pythagorean identity: Substitute the given value of cos(theta_(1)) into the equation. sin²(theta_(1)) = 1 - (-(12/19))² sin²(theta_(1)) = 1 - (144/361)Substituting the given value: Simplify the right side of the equation. sin²(theta_(1)) = (361/361) - (144/361) sin²(theta_(1)) = (361 - 144) / 361 sin²(theta_(1)) = 217 / 361Simplifying the equation: Take the square root of both sides to find sin(theta_(1)). Since theta_(1) is in Quadrant II, sin(theta_(1)) is positive. sin(theta_(1)) = sqrt(217 / 361) sin(theta_(1)) = sqrt(217) / 19

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The angle
theta_(1) is located in Quadrant II, and
cos(theta_(1))=-(12)/(19).
What is the value of
sin(theta_(1)) ? Express your answer exactly.

sin(theta_(1))=

The angle $\theta_{1}$ is located in Quadrant II, and $\cos \left(\theta_{1}\right)=-\frac{12}{19}$ . $\newline$ What is the value of $\sin \left(\theta_{1}\right)$ ? Express your answer exactly. $\newline$ $\sin \left(\theta_{1}\right)=$

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Q. The angle

\theta_{1}

is located in Quadrant II, and

\cos \left(\theta_{1}\right)=-\frac{12}{19}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Pythagorean identity: We know that for any angle $\theta$ , $\sin^2(\theta) + \cos^2(\theta) = 1$ (Pythagorean identity). $\newline$ Given that $\cos(\theta_{1}) = -\frac{12}{19}$ , we can find $\sin(\theta_{1})$ by rearranging the Pythagorean identity to solve for $\sin^2(\theta_{1})$ . $\newline$ $\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})$
Rearranging the Pythagorean identity: Substitute the given value of $\cos(\theta_{1})$ into the equation. $\sin^2(\theta_{1}) = 1 - (-\left(\frac{12}{19}\right))^2$ $\sin^2(\theta_{1}) = 1 - \left(\frac{144}{361}\right)$
Substituting the given value: Simplify the right side of the equation. $\newline$ $\sin^2(\theta_{1}) = \frac{361}{361} - \frac{144}{361}$ $\newline$ $\sin^2(\theta_{1}) = \frac{361 - 144}{361}$ $\newline$ $\sin^2(\theta_{1}) = \frac{217}{361}$
Simplifying the equation: Take the square root of both sides to find $\sin(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant II, $\sin(\theta_{1})$ is positive. $\newline$ $\sin(\theta_{1}) = \sqrt{\frac{217}{361}}$ $\newline$ $\sin(\theta_{1}) = \frac{\sqrt{217}}{19}$