The angle theta_(1) is located in Quadrant IV, and cos(theta_(1))=(9)/(19). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Use Pythagorean Identity: Use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to find sin(theta_(1)). cos(theta_(1)) = 9/19, so cos^2(theta_(1)) = (9/19)^2. Calculate cos^2(theta_(1)). cos^2(theta_(1)) = (9/19)^2 = 81/361.Calculate cos^2(theta_(1)): Substitute cos^2(theta_(1)) into the Pythagorean identity. sin^2(theta_(1)) + cos^2(theta_(1)) = 1 sin^2(theta_(1)) + 81/361 = 1 sin^2(theta_(1)) = 1 - 81/361 Calculate sin^2(theta_(1)). sin^2(theta_(1)) = 1 - 81/361 = 361/361 - 81/361 = 280/361Substitute into Identity: Find the value of sin(theta_(1)). Since theta_(1) is in Quadrant IV, sin(theta_(1)) is negative. Take the square root of sin^2(theta_(1)). sin(theta_(1)) = -sqrt(280/361) Simplify the square root. sin(theta_(1)) = -sqrt(280)/sqrt(361) sin(theta_(1)) = -sqrt(4*70)/19 sin(theta_(1)) = -2*sqrt(70)/19

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The angle
theta_(1) is located in Quadrant
IV, and
cos(theta_(1))=(9)/(19).
What is the value of
sin(theta_(1)) ?
Express your answer exactly.

sin(theta_(1))=

The angle $\theta_{1}$ is located in Quadrant $\newline$ IV, and $\cos \left(\theta_{1}\right)=\frac{9}{19}$ . $\newline$ What is the value of $\sin \left(\theta_{1}\right)$ ? $\newline$ Express your answer exactly. $\newline$ $\sin \left(\theta_{1}\right)=$

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Q. The angle

\theta_{1}

is located in Quadrant

\newline

IV, and

\cos \left(\theta_{1}\right)=\frac{9}{19}

\newline

What is the value of

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

\newline

Express your answer exactly.

\newline

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

Use Pythagorean Identity: Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find $\sin(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = \frac{9}{19}$ , so $\cos^2(\theta_{1}) = \left(\frac{9}{19}\right)^2$ . $\newline$ Calculate $\cos^2(\theta_{1})$ . $\newline$ $\cos^2(\theta_{1}) = \left(\frac{9}{19}\right)^2 = \frac{81}{361}$ .
Calculate $\cos^2(\theta_{1})$ : Substitute $\cos^2(\theta_{1})$ into the Pythagorean identity. $\newline$ $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ $\newline$ $\sin^2(\theta_{1}) + \frac{81}{361} = 1$ $\newline$ $\sin^2(\theta_{1}) = 1 - \frac{81}{361}$ $\newline$ Calculate $\sin^2(\theta_{1})$ . $\newline$ $\sin^2(\theta_{1}) = 1 - \frac{81}{361} = \frac{361}{361} - \frac{81}{361} = \frac{280}{361}$
Substitute into Identity: Find the value of $\sin(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant IV, $\sin(\theta_{1})$ is negative. Take the square root of $\sin^2(\theta_{1})$ . $\sin(\theta_{1}) = -\sqrt{\frac{280}{361}}$ Simplify the square root. $\sin(\theta_{1}) = -\frac{\sqrt{280}}{\sqrt{361}}$ $\sin(\theta_{1}) = -\frac{\sqrt{4*70}}{19}$ $\sin(\theta_{1}) = -\frac{2*\sqrt{70}}{19}$