The angle theta_(1) is located in Quadrant II, and sin(theta_(1))=(9)/(41). What is the value of cos(theta_(1)) ? Express your answer exactly. cos(theta_(1))=

Step 1: Find sin(theta_(1)): We know that sin(theta_(1)) = 9/41. Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find cos(theta_(1)). Substitute 9/41 for sin(theta_(1)) in sin²(theta_(1)) + cos²(theta_(1)) = 1. (9/41)² + cos²(theta_(1)) = 1.Step 2: Substitute sin(theta_(1)) in the Pythagorean identity: Simplify (9/41)² + cos²(theta_(1)) = 1 to find the value of cos(theta_(1)). (81/1681) + cos²(theta_(1)) = 1 cos²(theta_(1)) = 1 - 81/1681 cos²(theta_(1)) = (1681/1681) - (81/1681) cos²(theta_(1)) = 1600/1681Step 3: Simplify the equation: Since we are looking for cos(theta_(1)), we take the square root of both sides. cos(theta_(1)) = ±√(1600/1681) cos(theta_(1)) = ±(40/41)Step 4: Find the value of cos(theta_(1)): Determine the sign of cos(theta_(1)) based on the quadrant in which theta_(1) is located. Since theta_(1) is in Quadrant II, where cosine is negative, we choose the negative value. cos(theta_(1)) = -40/41

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Find trigonometric ratios using multiple identities

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Q. The angle

\theta_{1}

is located in Quadrant II, and

\sin \left(\theta_{1}\right)=\frac{9}{41}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Step $1$ : Find $\sin(\theta_{1})$ : We know that $\sin(\theta_{1}) = \frac{9}{41}$ . Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find $\cos(\theta_{1})$ . $\newline$ Substitute $\frac{9}{41}$ for $\sin(\theta_{1})$ in $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ . $\newline$ $\left(\frac{9}{41}\right)^2 + \cos^2(\theta_{1}) = 1$ .
Step $2$ : Substitute $\sin(\theta_{1})$ in the Pythagorean identity: Simplify $\left(\frac{9}{41}\right)^2 + \cos^2(\theta_{1}) = 1$ to find the value of $\cos(\theta_{1})$ . $\newline$ $\left(\frac{81}{1681}\right) + \cos^2(\theta_{1}) = 1$ $\newline$ $\cos^2(\theta_{1}) = 1 - \frac{81}{1681}$ $\newline$ $\cos^2(\theta_{1}) = \frac{1681}{1681} - \frac{81}{1681}$ $\newline$ $\cos^2(\theta_{1}) = \frac{1600}{1681}$
Step $3$ : Simplify the equation: Since we are looking for $\cos(\theta_{1})$ , we take the square root of both sides. $\newline$ $\cos(\theta_{1}) = \pm\sqrt{\frac{1600}{1681}}$ $\newline$ $\cos(\theta_{1}) = \pm\frac{40}{41}$
Step $4$ : Find the value of $\cos(\theta_{1})$ : Determine the sign of $\cos(\theta_{1})$ based on the quadrant in which $\theta_{1}$ is located. $\newline$ Since $\theta_{1}$ is in Quadrant II, where cosine is negative, we choose the negative value. $\newline$ $\cos(\theta_{1}) = -\frac{40}{41}$