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The angle 
theta_(1) is located in Quadrant III, and 
sin(theta_(1))=-(12)/(13).
What is the value of 
cos(theta_(1)) ?
Express your answer exactly.

cos(theta_(1))=

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The angle θ1 \theta_{1} is located in Quadrant III, and sin(θ1)=1213 \sin \left(\theta_{1}\right)=-\frac{12}{13} .\newlineWhat is the value of cos(θ1) \cos \left(\theta_{1}\right) ?\newlineExpress your answer exactly.\newlinecos(θ1)= \cos \left(\theta_{1}\right)= \newline \square

Full solution

Q. The angle θ1 \theta_{1} is located in Quadrant III, and sin(θ1)=1213 \sin \left(\theta_{1}\right)=-\frac{12}{13} .\newlineWhat is the value of cos(θ1) \cos \left(\theta_{1}\right) ?\newlineExpress your answer exactly.\newlinecos(θ1)= \cos \left(\theta_{1}\right)= \newline \square
  1. Apply Pythagorean Identity: Use the Pythagorean identity for sine and cosine.\newlineThe Pythagorean identity states that sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.\newlineSince we know sin(θ1)=1213\sin(\theta_{1}) = -\frac{12}{13}, we can substitute this into the identity to find cos(θ1)\cos(\theta_{1}).
  2. Substitute sin(θ1)\sin(\theta_{1}): Substitute the value of sin(θ1)\sin(\theta_{1}) into the Pythagorean identity.\newlinesin2(θ1)+cos2(θ1)=1\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1\newline(1213)2+cos2(θ1)=1(-\frac{12}{13})^2 + \cos^2(\theta_{1}) = 1\newline144169+cos2(θ1)=1\frac{144}{169} + \cos^2(\theta_{1}) = 1
  3. Solve for cos2(θ1)\cos^2(\theta_{1}): Solve for cos2(θ1)\cos^2(\theta_{1}).
    cos2(θ1)=1144169\cos^2(\theta_{1}) = 1 - \frac{144}{169}
    cos2(θ1)=169169144169\cos^2(\theta_{1}) = \frac{169}{169} - \frac{144}{169}
    cos2(θ1)=169144169\cos^2(\theta_{1}) = \frac{169 - 144}{169}
    cos2(θ1)=25169\cos^2(\theta_{1}) = \frac{25}{169}
  4. Find cos(θ1)\cos(\theta_{1}): Take the square root of both sides to find cos(θ1)\cos(\theta_{1}). Since θ1\theta_{1} is in Quadrant III, where cosine is negative, we take the negative square root. cos(θ1)=25169\cos(\theta_{1}) = -\sqrt{\frac{25}{169}} cos(θ1)=513\cos(\theta_{1}) = -\frac{5}{13}

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