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The angle
theta_(1) is located in Quadrant IV, and
cos(theta_(1))=(3)/(5).
What is the value of
sin(theta_(1)) ? Express your answer exactly.

sin(theta_(1))=

The angle $\theta_{1}$ is located in Quadrant IV, and $\cos \left(\theta_{1}\right)=\frac{3}{5}$ . $\newline$ What is the value of $\sin \left(\theta_{1}\right)$ ? Express your answer exactly. $\newline$ $\sin \left(\theta_{1}\right)=$

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Find trigonometric ratios using a Pythagorean or reciprocal identity

Full solution

Q. The angle

\theta_{1}

is located in Quadrant IV, and

\cos \left(\theta_{1}\right)=\frac{3}{5}

.

\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 $\cos(\theta_{1}) = \frac{3}{5}$ and that $\theta_{1}$ is in Quadrant IV. In Quadrant IV, cosine is positive and sine is negative. We can use the Pythagorean identity $\sin^{2}(\theta) + \cos^{2}(\theta) = 1$ to find the value of $\sin(\theta_{1})$ .
Solving for $\sin^2(\theta)$ : First, we will solve for $\sin^2(\theta_{1})$ . We know that $\cos^2(\theta_{1}) = \left(\frac{3}{5}\right)^2$ . $\cos^2(\theta_{1}) = \frac{9}{25}$
Using Pythagorean Identity: Now, we use the Pythagorean identity to find $\sin^2(\theta_{1})$ .
$\sin^2(\theta_{1}) = 1 - \cos^2(\theta_{1})$
$\sin^2(\theta_{1}) = 1 - \frac{9}{25}$
$\sin^2(\theta_{1}) = \frac{25}{25} - \frac{9}{25}$
$\sin^2(\theta_{1}) = \frac{16}{25}$
Finding $\sin(\theta)$ : Next, we take the square root of both sides to find $\sin(\theta_{1})$ . Since $\theta_{1}$ is in Quadrant IV, we must take the negative square root because sine is negative in this quadrant. $\newline$ $\sin(\theta_{1}) = -\sqrt{\frac{16}{25}}$ $\newline$ $\sin(\theta_{1}) = -\frac{4}{5}$

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