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The angle theta_(1) is located in Quadrant III, and cos(theta_(1))=-(5)/(8). 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)=−58 \cos \left(\theta_{1}\right)=-\frac{5}{8} .\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)=−58 \cos \left(\theta_{1}\right)=-\frac{5}{8} .\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: In Quadrant III, both sin⁡\sin and cos⁡\cos are negative. We are given that cos⁡(θ1)=−58\cos(\theta_{1}) = -\frac{5}{8}. To find sin⁡(θ1)\sin(\theta_{1}), we can use the Pythagorean identity: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\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)=(−(58))2=(58)2=2564.\cos^2(\theta_{1}) = (-(\frac{5}{8}))^2 = (\frac{5}{8})^2 = \frac{25}{64}.
  3. Finding sin⁡2(θ1)\sin^2(\theta_{1}): Next, we use the Pythagorean identity to find sin⁡2(θ1)\sin^2(\theta_{1}):\newlinesin⁡2(θ1)+cos⁡2(θ1)=1\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1\newlinesin⁡2(θ1)+2564=1\sin^2(\theta_{1}) + \frac{25}{64} = 1\newlinesin⁡2(θ1)=1−2564\sin^2(\theta_{1}) = 1 - \frac{25}{64}\newlinesin⁡2(θ1)=6464−2564\sin^2(\theta_{1}) = \frac{64}{64} - \frac{25}{64}\newlinesin⁡2(θ1)=3964\sin^2(\theta_{1}) = \frac{39}{64}.
  4. Finding sin⁡(θ1)\sin(\theta_{1}): Now, we take the square root of both sides to find sin⁡(θ1)\sin(\theta_{1}). Since we are in Quadrant III, where sine is negative, we take the negative square root:\newlinesin⁡(θ1)=−3964\sin(\theta_{1}) = -\sqrt{\frac{39}{64}}\newlinesin⁡(θ1)=−398\sin(\theta_{1}) = -\frac{\sqrt{39}}{8}.

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