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The angle theta_(1) is located in Quadrant III, and cos(theta_(1))=-(13)/(15). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

The angle θ1 \theta_{1} is located in Quadrant III, and cos⁡(θ1)=−1315 \cos \left(\theta_{1}\right)=-\frac{13}{15} .\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 III, and cos⁡(θ1)=−1315 \cos \left(\theta_{1}\right)=-\frac{13}{15} .\newlineWhat is the value of sin⁡(θ1) \sin \left(\theta_{1}\right) ? Express your answer exactly.\newlinesin⁡(θ1)= \sin \left(\theta_{1}\right)=
  1. Given information: We are given that cos⁡(θ1)=−1315\cos(\theta_{1}) = -\frac{13}{15} and θ1\theta_{1} is in Quadrant III. In Quadrant III, both sine and cosine are negative. We will use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to find the value of sin⁡(θ1)\sin(\theta_{1}).
  2. Finding cos⁡2(θ1)\cos^2(\theta_{1}): First, we square the given cosine value to find cos⁡2(θ1)\cos^2(\theta_{1}).\newlinecos⁡2(θ1)=(−1315)2\cos^2(\theta_{1}) = \left(-\frac{13}{15}\right)^2\newlinecos⁡2(θ1)=169225\cos^2(\theta_{1}) = \frac{169}{225}
  3. Finding sin⁡2(θ1)\sin^2(\theta_{1}): Next, we use the Pythagorean identity to find sin⁡2(θ1)\sin^2(\theta_{1}).
    sin⁡2(θ1)=1−cos⁡2(θ1)\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})
    sin⁡2(θ1)=1−169225\sin^2(\theta_{1}) = 1 - \frac{169}{225}
    sin⁡2(θ1)=225225−169225\sin^2(\theta_{1}) = \frac{225}{225} - \frac{169}{225}
    sin⁡2(θ1)=56225\sin^2(\theta_{1}) = \frac{56}{225}
  4. Finding sin⁡(θ1)\sin(\theta_{1}): Now, we take the square root of both sides to find sin⁡(θ1)\sin(\theta_{1}). Since θ1\theta_{1} is in Quadrant III, where sine is negative, we take the negative square root.\newlinesin⁡(θ1)=−56225\sin(\theta_{1}) = -\sqrt{\frac{56}{225}}\newlinesin⁡(θ1)=−56225\sin(\theta_{1}) = -\frac{\sqrt{56}}{\sqrt{225}}\newlinesin⁡(θ1)=−4⋅1415\sin(\theta_{1}) = -\frac{\sqrt{4 \cdot 14}}{15}\newlinesin⁡(θ1)=−2⋅1415\sin(\theta_{1}) = -\frac{2 \cdot \sqrt{14}}{15}

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