The angle theta_(1) is located in Quadrant III, and cos(theta_(1))=-(5)/(8). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Given information: In Quadrant III, both sine and cosine are negative. We are given that cos(theta_(1)) = -(5/8). To find sin(theta_(1)), we can use the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1.Finding cos^2(theta_(1)): First, we square the given cosine value to find cos^2(theta_(1)): cos^2(theta_(1)) = (-(5/8))^2 = (5/8)^2 = 25/64.Finding sin^2(theta_(1)): Next, we use the Pythagorean identity to find sin^2(theta_(1)): sin^2(theta_(1)) + cos^2(theta_(1)) = 1 sin^2(theta_(1)) + 25/64 = 1 sin^2(theta_(1)) = 1 - 25/64 sin^2(theta_(1)) = (64/64) - (25/64) sin^2(theta_(1)) = 39/64.Finding sin(theta_(1)): Now, we take the square root of both sides to find sin(theta_(1)). Since we are in Quadrant III, where sine is negative, we take the negative square root: sin(theta_(1)) = -sqrt(39/64) sin(theta_(1)) = -sqrt(39)/8.

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Find trigonometric ratios using a Pythagorean or reciprocal identity

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Q. The angle

\theta_{1}

is located in Quadrant III, and

\cos \left(\theta_{1}\right)=-\frac{5}{8}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Given information: In Quadrant III, both $\sin$ and $\cos$ are negative. We are given that $\cos(\theta_{1}) = -\frac{5}{8}$ . To find $\sin(\theta_{1})$ , we can use the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ .
Finding $\cos^2(\theta_{1})$ : First, we square the given cosine value to find $\cos^2(\theta_{1})$ : $\newline$ $\cos^2(\theta_{1}) = (-(\frac{5}{8}))^2 = (\frac{5}{8})^2 = \frac{25}{64}.$
Finding $\sin^2(\theta_{1})$ : Next, we use the Pythagorean identity to find $\sin^2(\theta_{1})$ : $\newline$ $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ $\newline$ $\sin^2(\theta_{1}) + \frac{25}{64} = 1$ $\newline$ $\sin^2(\theta_{1}) = 1 - \frac{25}{64}$ $\newline$ $\sin^2(\theta_{1}) = \frac{64}{64} - \frac{25}{64}$ $\newline$ $\sin^2(\theta_{1}) = \frac{39}{64}$ .
Finding $\sin(\theta_{1})$ : Now, we take the square root of both sides to find $\sin(\theta_{1})$ . Since we are in Quadrant III, where sine is negative, we take the negative square root: $\newline$ $\sin(\theta_{1}) = -\sqrt{\frac{39}{64}}$ $\newline$ $\sin(\theta_{1}) = -\frac{\sqrt{39}}{8}$ .