The angle theta_(1) is located in Quadrant II, and sin(theta_(1))=(1)/(4). What is the value of cos(theta_(1)) ? Express your answer exactly. cos(theta_(1))=

Use Pythagorean Identity: We know that sin(theta_(1)) = 1/4. Use the Pythagorean identity sin²(theta) + cos²(theta) = 1 to find the value of cos(theta_(1)). Substitute 1/4 for sin(theta_(1)) in sin²(theta_(1)) + cos²(theta_(1)) = 1. (1/4)² + cos²(theta_(1)) = 1.Substitute and Simplify: Simplify (1/4)² + cos²(theta_(1)) = 1 to find the value of cos²(theta_(1)). (1/16) + cos²(theta_(1)) = 1 cos²(theta_(1)) = 1 - 1/16 cos²(theta_(1)) = 15/16Find cos(theta_(1)): Since cos²(theta_(1)) = 15/16, we take the square root of both sides to find cos(theta_(1)). cos(theta_(1)) = ±√(15/16) cos(theta_(1)) = ±(√15)/4Determine Sign: Determine the sign of cos(theta_(1)) based on the quadrant in which theta_(1) is located. Since theta_(1) is in Quadrant II, where cosine is negative, we choose the negative value for cos(theta_(1)). cos(theta_(1)) = -(√15)/4

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The angle
theta_(1) is located in Quadrant II, and
sin(theta_(1))=(1)/(4).
What is the value of
cos(theta_(1)) ? Express your answer exactly.

cos(theta_(1))=

The angle $\theta_{1}$ is located in Quadrant II, and $\sin \left(\theta_{1}\right)=\frac{1}{4}$ . $\newline$ What is the value of $\cos \left(\theta_{1}\right)$ ? Express your answer exactly. $\newline$ $\cos \left(\theta_{1}\right)=$

View step-by-step help

Home

Math Problems

Precalculus

Find trigonometric ratios using multiple identities

Full solution

Q. The angle

\theta_{1}

is located in Quadrant II, and

\sin \left(\theta_{1}\right)=\frac{1}{4}

\newline

What is the value of

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

? Express your answer exactly.

\newline

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

Use Pythagorean Identity: We know that $\sin(\theta_{1}) = \frac{1}{4}$ . Use the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$ to find the value of $\cos(\theta_{1})$ . Substitute $\frac{1}{4}$ for $\sin(\theta_{1})$ in $\sin^2(\theta_{1}) + \cos^2(\theta_{1}) = 1$ . $(\frac{1}{4})^2 + \cos^2(\theta_{1}) = 1$ .
Substitute and Simplify: Simplify $(\frac{1}{4})^2 + \cos^2(\theta_{1}) = 1$ to find the value of $\cos^2(\theta_{1})$ . $\frac{1}{16} + \cos^2(\theta_{1}) = 1$ $\cos^2(\theta_{1}) = 1 - \frac{1}{16}$ $\cos^2(\theta_{1}) = \frac{15}{16}$
Find $\cos(\theta_{1})$ : Since $\cos^2(\theta_{1}) = \frac{15}{16}$ , we take the square root of both sides to find $\cos(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = \pm\sqrt{\frac{15}{16}}$ $\newline$ $\cos(\theta_{1}) = \pm\frac{\sqrt{15}}{4}$
Determine Sign: Determine the sign of $\cos(\theta_{1})$ based on the quadrant in which $\theta_{1}$ is located. $\newline$ Since $\theta_{1}$ is in Quadrant II, where cosine is negative, we choose the negative value for $\cos(\theta_{1})$ . $\newline$ $\cos(\theta_{1}) = -\left(\sqrt{15}\right)/4$