The angle theta_(1) is located in Quadrant II, and cos(theta_(1))=-(2)/(11). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Use Pythagorean Identity: We know that cos(theta_(1)) = -(2/11). Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find sin(theta_(1)). Substitute -(2/11) for cos(theta_(1)) in the identity. sin²(theta_(1)) + (-(2/11))² = 1.Simplify to Find sin: Simplify the equation to find the value of sin²(theta_(1)). sin²(theta_(1)) + (4/121) = 1. sin²(theta_(1)) = 1 - (4/121). sin²(theta_(1)) = (121/121) - (4/121). sin²(theta_(1)) = (117/121).Find sin(theta): Find the square root of sin²(theta_(1)) to get sin(theta_(1)). Since theta_(1) is in Quadrant II, sin(theta_(1)) is positive. sin(theta_(1)) = sqrt(117/121). sin(theta_(1)) = sqrt(117)/sqrt(121). sin(theta_(1)) = sqrt(117)/11.Determine Simplified Form: Determine the simplified form of sqrt(117). 117 is 3 * 3 * 13, so sqrt(117) simplifies to 3 * sqrt(13). Therefore, sin(theta_(1)) = (3 * sqrt(13))/11.

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The angle
theta_(1) is located in Quadrant II, and
cos(theta_(1))=-(2)/(11).
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 $\cos \left(\theta_{1}\right)=-\frac{2}{11}$ . $\newline$ 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{2}{11}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Use Pythagorean Identity: We know that $\cos(\theta_{1}) = -\frac{2}{11}$ . Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find $\sin(\theta_{1})$ . $\newline$ Substitute $-\frac{2}{11}$ for $\cos(\theta_{1})$ in the identity. $\newline$ $\sin^2(\theta_{1}) + \left(-\frac{2}{11}\right)^2 = 1$ .
Simplify to Find sin: Simplify the equation to find the value of $\sin^2(\theta_{1})$ . $\sin^2(\theta_{1}) + \frac{4}{121} = 1$ . $\sin^2(\theta_{1}) = 1 - \frac{4}{121}$ . $\sin^2(\theta_{1}) = \frac{121}{121} - \frac{4}{121}$ . $\sin^2(\theta_{1}) = \frac{117}{121}$ .
Find $\sin(\theta):$ Find the square root of $\sin^2(\theta_{1})$ to get $\sin(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant II, $\sin(\theta_{1})$ is positive. $\sin(\theta_{1}) = \sqrt{117/121}$ . $\sin(\theta_{1}) = \sqrt{117}/\sqrt{121}$ . $\sin(\theta_{1}) = \sqrt{117}/11$ .
Determine Simplified Form: Determine the simplified form of $\sqrt{117}$ . $117$ is $3 \times 3 \times 13$ , so $\sqrt{117}$ simplifies to $3 \times \sqrt{13}$ . Therefore, $\sin(\theta_{1}) = \frac{3 \times \sqrt{13}}{11}$ .