The angle theta_(1) is located in Quadrant I, and sin(theta_(1))=(11)/(61). What is the value of cos(theta_(1)) ? Express your answer exactly. cos(theta_(1))=

Use Pythagorean Identity: We know that sin(theta_(1)) = 11/61. Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find the value of cos(theta_(1)). Substitute 11/61 for sin(theta_(1)) in sin²(theta) + cos²(theta) = 1. (11/61)² + cos²(theta_(1)) = 1.Substitute sin(theta): Simplify (11/61)² + cos²(theta_(1)) = 1 to find the value of cos²(theta_(1)). (11/61)² = 121/3721. cos²(theta_(1)) = 1 - 121/3721.Simplify to find cos(theta): Calculate cos²(theta_(1)) = 1 - 121/3721. cos²(theta_(1)) = 3721/3721 - 121/3721. cos²(theta_(1)) = (3721 - 121) / 3721. cos²(theta_(1)) = 3600 / 3721.Calculate cos(theta): Since theta_(1) is in Quadrant I, where all trigonometric functions are positive, we take the positive square root of cos²(theta_(1)). cos(theta_(1)) = sqrt(3600 / 3721). cos(theta_(1)) = 60/61.

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The angle
theta_(1) is located in Quadrant
I, and
sin(theta_(1))=(11)/(61).
What is the value of
cos(theta_(1)) ?
Express your answer exactly.

cos(theta_(1))=

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

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Q. The angle

\theta_{1}

is located in Quadrant

\newline

I, and

\sin \left(\theta_{1}\right)=\frac{11}{61}

\newline

What is the value of

\cos \left(\theta_{1}\right)

\newline

Express your answer exactly.

\newline

\cos \left(\theta_{1}\right)=

Use Pythagorean Identity: We know that $\sin(\theta_{1}) = \frac{11}{61}$ . Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find the value of $\cos(\theta_{1})$ . $\newline$ Substitute $\frac{11}{61}$ for $\sin(\theta_{1})$ in $\sin^2(\theta) + \cos^2(\theta) = 1$ . $\newline$ $(\frac{11}{61})^2 + \cos^2(\theta_{1}) = 1$ .
Substitute $\sin(\theta)$ : Simplify $\left(\frac{11}{61}\right)^2 + \cos^2(\theta_{1}) = 1$ to find the value of $\cos^2(\theta_{1})$ . $\left(\frac{11}{61}\right)^2 = \frac{121}{3721}$ . $\cos^2(\theta_{1}) = 1 - \frac{121}{3721}$ .
Simplify to find $\cos(\theta)$ : Calculate $\cos^2(\theta_{1}) = 1 - \frac{121}{3721}$ .
$\cos^2(\theta_{1}) = \frac{3721}{3721} - \frac{121}{3721}$ .
$\cos^2(\theta_{1}) = \frac{3721 - 121}{3721}$ .
$\cos^2(\theta_{1}) = \frac{3600}{3721}$ .
Calculate $\cos(\theta)$ : Since $\theta_{1}$ is in Quadrant I, where all trigonometric functions are positive, we take the positive square root of $\cos^2(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = \sqrt{\frac{3600}{3721}}$ . $\newline$ $\cos(\theta_{1}) = \frac{60}{61}$ .