The angle theta_(1) is located in Quadrant IV, and sin(theta_(1))=-(10)/(13). What is the value of cos(theta_(1)) ? Express your answer exactly. cos(theta_(1))=

Step 1: Find sin(theta_(1)): We know that sin(theta_(1)) = -(10)/(13). Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find the value of cos(theta_(1)). Substitute -(10)/(13) for sin(theta_(1)) in sin²(theta_(1)) + cos²(theta_(1)) = 1. (-10/13)² + cos²(theta_(1)) = 1.Step 2: Use Pythagorean identity: Simplify (-10/13)² + cos²(theta_(1)) = 1 to find the value of cos²(theta_(1)). (100/169) + cos²(theta_(1)) = 1 cos²(theta_(1)) = 1 - 100/169 cos²(theta_(1)) = 169/169 - 100/169 cos²(theta_(1)) = 69/169Step 3: Substitute sin(theta_(1)): Find the square root of cos²(theta_(1)) to get cos(theta_(1)). cos(theta_(1)) = ±√(69/169) cos(theta_(1)) = ±√69/13Step 4: Simplify the equation: Determine the sign of cos(theta_(1)) based on the quadrant in which theta_(1) is located. Since theta_(1) is in Quadrant IV, where cosine is positive, we choose the positive square root. cos(theta_(1)) = √69/13

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Find trigonometric ratios using multiple identities

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Q. The angle

\theta_{1}

is located in Quadrant

\newline

IV, and

\sin \left(\theta_{1}\right)=-\frac{10}{13}

\newline

What is the value of

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

\newline

Express your answer exactly.

\newline

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

Step $1$ : Find $\sin(\theta_{1})$ : We know that $\sin(\theta_{1}) = -\frac{10}{13}$ . $\newline$ Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find the value of $\cos(\theta_{1})$ . $\newline$ Substitute $-\frac{10}{13}$ for $\sin(\theta_{1})$ in $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ . $\newline$ $\left(-\frac{10}{13}\right)^2 + \cos^2(\theta_{1}) = 1$ .
Step $2$ : Use Pythagorean identity: Simplify $(-\frac{10}{13})^2 + \cos^2(\theta_{1}) = 1$ to find the value of $\cos^2(\theta_{1})$ . $\newline$ $\frac{100}{169} + \cos^2(\theta_{1}) = 1$ $\newline$ $\cos^2(\theta_{1}) = 1 - \frac{100}{169}$ $\newline$ $\cos^2(\theta_{1}) = \frac{169}{169} - \frac{100}{169}$ $\newline$ $\cos^2(\theta_{1}) = \frac{69}{169}$
Step $3$ : Substitute $\sin(\theta_{1})$ : Find the square root of $\cos^2(\theta_{1})$ to get $\cos(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = \pm\sqrt{\frac{69}{169}}$ $\newline$ $\cos(\theta_{1}) = \pm\frac{\sqrt{69}}{13}$
Step $4$ : Simplify the equation: Determine the sign of $\cos(\theta_{1})$ based on the quadrant in which $\theta_{1}$ is located. $\newline$ Since $\theta_{1}$ is in Quadrant IV, where cosine is positive, we choose the positive square root. $\newline$ $\cos(\theta_{1}) = \frac{\sqrt{69}}{13}$