The angle theta_(1) is located in Quadrant I, and cos(theta_(1))=(10)/(17). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Use Pythagorean Identity: We know that cos(theta_(1)) = 10/17. To find sin(theta_(1)), we use the Pythagorean identity sin²(theta_(1)) + cos²(theta_(1)) = 1. Substitute cos(theta_(1)) = 10/17 into the identity. sin²(theta_(1)) + (10/17)² = 1.Simplify and Calculate sin(theta_(1)): Simplify (10/17)² and subtract it from 1 to find sin²(theta_(1)). sin²(theta_(1)) + 100/289 = 1. sin²(theta_(1)) = 1 - 100/289. sin²(theta_(1)) = 289/289 - 100/289. sin²(theta_(1)) = 189/289.Take Square Root: Take the square root of both sides to find sin(theta_(1)). sin(theta_(1)) = ±√(189/289). Since theta_(1) is in Quadrant I, where sine is positive, we choose the positive root. sin(theta_(1)) = √(189/289).Final Simplification: Simplify the square root of the fraction. sin(theta_(1)) = √189 / √289. sin(theta_(1)) = √(9*21) / 17. sin(theta_(1)) = 3√21 / 17.

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The angle
theta_(1) is located in Quadrant
I, and
cos(theta_(1))=(10)/(17).
What is the value of
sin(theta_(1)) ?
Express your answer exactly.

sin(theta_(1))=

The angle $\theta_{1}$ is located in Quadrant $\newline$ I, and $\cos \left(\theta_{1}\right)=\frac{10}{17}$ . $\newline$ What is the value of $\sin \left(\theta_{1}\right)$ ? $\newline$ Express your answer exactly. $\newline$ $\sin \left(\theta_{1}\right)=$

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Q. The angle

\theta_{1}

is located in Quadrant

\newline

I, and

\cos \left(\theta_{1}\right)=\frac{10}{17}

\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 $\cos(\theta_{1}) = \frac{10}{17}$ . To find $\sin(\theta_{1})$ , we use the Pythagorean identity $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ . $\newline$ Substitute $\cos(\theta_{1}) = \frac{10}{17}$ into the identity. $\newline$ $\sin^2(\theta_{1}) + \left(\frac{10}{17}\right)^2 = 1$ .
Simplify and Calculate $\sin(\theta_{1})$ : Simplify $(\frac{10}{17})^2$ and subtract it from $1$ to find $\sin^2(\theta_{1})$ .
$\sin^2(\theta_{1}) + \frac{100}{289} = 1$ .
$\sin^2(\theta_{1}) = 1 - \frac{100}{289}$ .
$\sin^2(\theta_{1}) = \frac{289}{289} - \frac{100}{289}$ .
$\sin^2(\theta_{1}) = \frac{189}{289}$ .
Take Square Root: Take the square root of both sides to find $\sin(\theta_{1})$ . $\sin(\theta_{1}) = \pm\sqrt{\frac{189}{289}}$ .Since $\theta_{1}$ is in Quadrant I, where sine is positive, we choose the positive root. $\sin(\theta_{1}) = \sqrt{\frac{189}{289}}$ .
Final Simplification: Simplify the square root of the fraction. $\newline$ $\sin(\theta_{1}) = \frac{\sqrt{189}}{\sqrt{289}}.$ $\newline$ $\sin(\theta_{1}) = \frac{\sqrt{9\cdot21}}{17}.$ $\newline$ $\sin(\theta_{1}) = \frac{3\sqrt{21}}{17}.$