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Posted 3 months ago

The angle θ1\theta_{1} is located in Quadrant I\mathrm{I}, and sin(θ1)=1161\sin \left(\theta_{1}\right)=\frac{11}{61}.
What is the value of cos(θ1)\cos \left(\theta_{1}\right) ?
Express your answer exactly.
\cos \left(\theta_{1}\right)=

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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Given that sin(θ1)=1161\sin(\theta_1) = \frac{11}{61} and θ1\theta_1 is in Quadrant I
step 2
Use the Pythagorean identity: sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1
step 3
Substitute sin(θ1)\sin(\theta_1) into the identity: (1161)2+cos2(θ1)=1\left(\frac{11}{61}\right)^2 + \cos^2(\theta_1) = 1
step 4
Calculate sin2(θ1)\sin^2(\theta_1): (1161)2=1213721\left(\frac{11}{61}\right)^2 = \frac{121}{3721}
step 5
Solve for cos2(θ1)\cos^2(\theta_1): cos2(θ1)=11213721\cos^2(\theta_1) = 1 - \frac{121}{3721}
step 6
Simplify the expression: cos2(θ1)=372137211213721=36003721\cos^2(\theta_1) = \frac{3721}{3721} - \frac{121}{3721} = \frac{3600}{3721}
step 7
Take the square root of cos2(θ1)\cos^2(\theta_1) to find cos(θ1)\cos(\theta_1): cos(θ1)=36003721\cos(\theta_1) = \sqrt{\frac{3600}{3721}}
step 8
Since θ1\theta_1 is in Quadrant I, cos(θ1)\cos(\theta_1) is positive: cos(θ1)=6061\cos(\theta_1) = \frac{60}{61}
[1] Answer
cos(θ1)=6061\cos(\theta_1) = \frac{60}{61}
Key Concept
Pythagorean Identity in Trigonometry
The Pythagorean identity sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 is used to find the cosine of an angle when the sine is known, especially when the angle is in the first quadrant where both sine and cosine are positive.

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