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The angle 
theta_(1) is located in Quadrant I, and 
sin(theta_(1))=(17)/(20).
What is the value of 
cos(theta_(1)) ?
Express your answer exactly.

The angle θ1 \theta_{1} is located in Quadrant I, and sin(θ1)=1720 \sin \left(\theta_{1}\right)=\frac{17}{20} .\newlineWhat is the value of cos(θ1) \cos \left(\theta_{1}\right) ?\newlineExpress your answer exactly.

Full solution

Q. The angle θ1 \theta_{1} is located in Quadrant I, and sin(θ1)=1720 \sin \left(\theta_{1}\right)=\frac{17}{20} .\newlineWhat is the value of cos(θ1) \cos \left(\theta_{1}\right) ?\newlineExpress your answer exactly.
  1. Apply Pythagorean Identity: Use the Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to find the value of cos(θ1)\cos(\theta_{1}). \newlinesin2(θ1)+cos2(θ1)=1\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1\newline(1720)2+cos2(θ1)=1(\frac{17}{20})^2 + \cos^2(\theta_{1}) = 1
  2. Calculate (17/20)2(17/20)^2: Calculate (17/20)2(17/20)^2.(17/20)2=(172)/(202)=289/400(17/20)^2 = (17^2)/(20^2) = 289/400
  3. Substitute into Identity: Substitute the value of (17/20)2(17/20)^2 into the Pythagorean identity.\newline289/400+cos2(θ1)=1289/400 + \cos^2(\theta_{1}) = 1
  4. Solve for cos2(θ1)\cos^2(\theta_{1}): Solve for cos2(θ1)\cos^2(\theta_{1}).\newlinecos2(θ1)=1289400\cos^2(\theta_{1}) = 1 - \frac{289}{400}\newlinecos2(θ1)=400400289400\cos^2(\theta_{1}) = \frac{400}{400} - \frac{289}{400}\newlinecos2(θ1)=400289400\cos^2(\theta_{1}) = \frac{400 - 289}{400}\newlinecos2(θ1)=111400\cos^2(\theta_{1}) = \frac{111}{400}
  5. Find cos(θ1)\cos(\theta_{1}): Find the square root of cos2(θ1)\cos^2(\theta_{1}) to get cos(θ1)\cos(\theta_{1}). Since θ1\theta_{1} is in Quadrant I, cos(θ1)\cos(\theta_{1}) is positive. cos(θ1)=111400\cos(\theta_{1}) = \sqrt{\frac{111}{400}} cos(θ1)=111400\cos(\theta_{1}) = \frac{\sqrt{111}}{\sqrt{400}} cos(θ1)=11120\cos(\theta_{1}) = \frac{\sqrt{111}}{20}

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