The angle theta_(1) is located in Quadrant III, and cos(theta_(1))=-(13)/(15). What is the value of sin(theta_(1)) ? Express your answer exactly. sin(theta_(1))=

Given information: We are given that cos(theta_(1)) = -(13)/(15) and theta_(1) is in Quadrant III. In Quadrant III, both sine and cosine are negative. We will use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to find the value of sin(theta_(1)).Finding cos^2(theta_(1)): First, we square the given cosine value to find cos^2(theta_(1)). cos^2(theta_(1)) = (-(13)/(15))^2 cos^2(theta_(1)) = (169)/(225)Finding sin^2(theta_(1)): Next, we use the Pythagorean identity to find sin^2(theta_(1)). sin^2(theta_(1)) = 1 - cos^2(theta_(1)) sin^2(theta_(1)) = 1 - (169)/(225) sin^2(theta_(1)) = (225)/(225) - (169)/(225) sin^2(theta_(1)) = (56)/(225)Finding sin(theta_(1)): Now, we take the square root of both sides to find sin(theta_(1)). Since theta_(1) is in Quadrant III, where sine is negative, we take the negative square root. sin(theta_(1)) = -sqrt((56)/(225)) sin(theta_(1)) = -sqrt(56)/sqrt(225) sin(theta_(1)) = -sqrt(4 * 14)/15 sin(theta_(1)) = -(2 * sqrt(14))/15

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Algebra 2

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{13}{15}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Given information: We are given that $\cos(\theta_{1}) = -\frac{13}{15}$ and $\theta_{1}$ is in Quadrant III. In Quadrant III, both sine and cosine are negative. We will use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find the value of $\sin(\theta_{1})$ .
Finding $\cos^2(\theta_{1})$ : First, we square the given cosine value to find $\cos^2(\theta_{1})$ . $\newline$ $\cos^2(\theta_{1}) = \left(-\frac{13}{15}\right)^2$ $\newline$ $\cos^2(\theta_{1}) = \frac{169}{225}$
Finding $\sin^2(\theta_{1})$ : Next, we use the Pythagorean identity to find $\sin^2(\theta_{1})$ .
$\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})$
$\sin^2(\theta_{1}) = 1 - \frac{169}{225}$
$\sin^2(\theta_{1}) = \frac{225}{225} - \frac{169}{225}$
$\sin^2(\theta_{1}) = \frac{56}{225}$
Finding $\sin(\theta_{1})$ : Now, we take the square root of both sides to find $\sin(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant III, where sine is negative, we take the negative square root. $\newline$ $\sin(\theta_{1}) = -\sqrt{\frac{56}{225}}$ $\newline$ $\sin(\theta_{1}) = -\frac{\sqrt{56}}{\sqrt{225}}$ $\newline$ $\sin(\theta_{1}) = -\frac{\sqrt{4 \cdot 14}}{15}$ $\newline$ $\sin(\theta_{1}) = -\frac{2 \cdot \sqrt{14}}{15}$