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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 θ1 \theta_{1} is located in Quadrant II, and cos⁡(θ1)=−1219 \cos \left(\theta_{1}\right)=-\frac{12}{19} .\newlineWhat is the value of sin⁡(θ1) \sin \left(\theta_{1}\right) ? Express your answer exactly.\newlinesin⁡(θ1)= \sin \left(\theta_{1}\right)=

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Q. The angle θ1 \theta_{1} is located in Quadrant II, and cos⁡(θ1)=−1219 \cos \left(\theta_{1}\right)=-\frac{12}{19} .\newlineWhat is the value of sin⁡(θ1) \sin \left(\theta_{1}\right) ? Express your answer exactly.\newlinesin⁡(θ1)= \sin \left(\theta_{1}\right)=
  1. Pythagorean identity: We know that for any angle θ\theta, sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 (Pythagorean identity).\newlineGiven that cos⁡(θ1)=−1219\cos(\theta_{1}) = -\frac{12}{19}, we can find sin⁡(θ1)\sin(\theta_{1}) by rearranging the Pythagorean identity to solve for sin⁡2(θ1)\sin^2(\theta_{1}).\newlinesin⁡2(θ1)=1−cos⁡2(θ1)\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})
  2. Rearranging the Pythagorean identity: Substitute the given value of cos⁡(θ1) \cos(\theta_{1}) into the equation.sin⁡2(θ1)=1−(−(1219))2\sin^2(\theta_{1}) = 1 - (-\left(\frac{12}{19}\right))^2sin⁡2(θ1)=1−(144361)\sin^2(\theta_{1}) = 1 - \left(\frac{144}{361}\right)
  3. Substituting the given value: Simplify the right side of the equation.\newlinesin⁡2(θ1)=361361−144361\sin^2(\theta_{1}) = \frac{361}{361} - \frac{144}{361}\newlinesin⁡2(θ1)=361−144361\sin^2(\theta_{1}) = \frac{361 - 144}{361}\newlinesin⁡2(θ1)=217361\sin^2(\theta_{1}) = \frac{217}{361}
  4. Simplifying the equation: Take the square root of both sides to find sin⁡(θ1)\sin(\theta_{1}). Since θ1\theta_{1} is in Quadrant II, sin⁡(θ1)\sin(\theta_{1}) is positive.\newlinesin⁡(θ1)=217361\sin(\theta_{1}) = \sqrt{\frac{217}{361}}\newlinesin⁡(θ1)=21719\sin(\theta_{1}) = \frac{\sqrt{217}}{19}

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