Use the given information to find the exact value of each of the following: cos⁡2θ\cos2\thetasin⁡θ=1213,θ lies in quadrant II.\sin \theta = \frac{12}{13}, \quad \theta \text{ lies in quadrant II.}

Ch. 3 - Trigonometric Identities and Equations

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Blitzer 3rd Edition

Ch. 3 - Trigonometric Identities and Equations

문제 8b

3장, 문제 8b

Use the given information to find the exact value of each of the following: $cosine 2 theta$
$\sin θ = \frac{12}{13},$

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Identify the given information: $\sin \theta = \frac{12}{13}$ and $\theta$ lies in quadrant II.

Recall the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. Use this to find $\cos \theta$.

Calculate $\cos \theta$ by rearranging the identity: $\cos \theta = \pm \sqrt{1 - \sin^2 \theta} = \pm \sqrt{1 - \left(\frac{12}{13}\right)^2}$.

Determine the correct sign of $\cos \theta$ based on the quadrant. Since $\theta$ is in quadrant II, $\cos \theta$ is negative.

Use the double-angle formula for cosine: $\cos 2\theta = 2 \cos^2 \theta - 1$. Substitute the value of $\cos \theta$ found in the previous step to express $\cos 2\theta$.

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주요 개념

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Pythagorean Identity

The Pythagorean identity states that sin²θ + cos²θ = 1 for any angle θ. Given sin θ, this identity allows you to find cos θ by rearranging the equation to cos²θ = 1 - sin²θ. This is essential for determining the cosine value when only sine is known.

Sign of Trigonometric Functions in Quadrants

The sign of sine and cosine depends on the quadrant where the angle lies. In quadrant II, sine is positive and cosine is negative. This knowledge helps assign the correct sign to cos θ after calculating its magnitude using the Pythagorean identity.

Double-Angle Formula for Cosine

The double-angle formula for cosine is cos 2θ = 2 cos²θ - 1 or cos 2θ = 1 - 2 sin²θ. This formula allows you to find cos 2θ using either sine or cosine of θ, making it crucial for solving the problem when sin θ is given.

Use the given information to find the exact value of each of the following: cos2θ\(\cos\)2\(\theta\)sinθ=1213,θ lies in quadrant II.\(\sin\) \(\theta\) = \(\frac{12}{13}\), \(\quad\) \(\theta\) \(\text{ lies in quadrant II.}\)