The angle theta_(1) is located in Quadrant I, and cos(theta_(1))=(3)/(8). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Use Pythagorean Identity: We know that cos(theta_(1)) = 3/8 and theta_(1) is in Quadrant I. Use the Pythagorean identity sin²(theta_(1)) + cos²(theta_(1)) = 1 to find sin(theta_(1)). Substitute 3/8 for cos(theta_(1)) in the identity. sin²(theta_(1)) + (3/8)² = 1.Simplify to Find sin(theta): Simplify (3/8)² + sin²(theta_(1)) = 1 to find the value of sin²(theta_(1)). (3/8)² = 9/64. sin²(theta_(1)) = 1 - 9/64. sin²(theta_(1)) = (64/64) - (9/64). sin²(theta_(1)) = 55/64.Find sin(theta): Find the square root of sin²(theta_(1)) to get sin(theta_(1)). sin(theta_(1)) = ±√(55/64). sin(theta_(1)) = ±√55/8.Determine Sign of sin(theta): Determine the sign of sin(theta_(1)) based on the quadrant where theta_(1) is located. Since theta_(1) is in Quadrant I, where both sine and cosine are positive, sin(theta_(1)) will be positive. sin(theta_(1)) = √55/8.

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Q. The angle

\theta_{1}

is located in Quadrant

\newline

I, and

\cos \left(\theta_{1}\right)=\frac{3}{8}

\newline

What is the value of

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

\newline

Express your answer exactly.

\newline

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

Use Pythagorean Identity: We know that $\cos(\theta_{1}) = \frac{3}{8}$ and $\theta_{1}$ is in Quadrant I. Use the Pythagorean identity $\sin^{2}(\theta_{1}) + \cos^{2}(\theta_{1}) = 1$ to find $\sin(\theta_{1})$ . $\newline$ Substitute $\frac{3}{8}$ for $\cos(\theta_{1})$ in the identity. $\newline$ $\sin^{2}(\theta_{1}) + \left(\frac{3}{8}\right)^{2} = 1$ .
Simplify to Find $\sin(\theta):$ Simplify $(\frac{3}{8})^2 + \sin^2(\theta_{1}) = 1$ to find the value of $\sin^2(\theta_{1})$ . $\left(\frac{3}{8}\right)^2 = \frac{9}{64}$ . $\sin^2(\theta_{1}) = 1 - \frac{9}{64}$ . $\sin^2(\theta_{1}) = \left(\frac{64}{64}\right) - \left(\frac{9}{64}\right)$ . $\sin^2(\theta_{1}) = \frac{55}{64}$ .
Find $\sin(\theta):$ Find the square root of $\sin^2(\theta_{1})$ to get $\sin(\theta_{1})$ . $\newline$ $\sin(\theta_{1}) = \pm\sqrt{\frac{55}{64}}.$ $\newline$ $\sin(\theta_{1}) = \pm\sqrt{55}/8.$
Determine Sign of $\sin(\theta)$ : Determine the sign of $\sin(\theta_{1})$ based on the quadrant where $\theta_{1}$ is located. $\newline$ Since $\theta_{1}$ is in Quadrant I, where both sine and cosine are positive, $\sin(\theta_{1})$ will be positive. $\newline$ $\sin(\theta_{1}) = \sqrt{\frac{55}{8}}.$