Questions: Andrea Myers nework #18: Section 6.3 (10 questions) Question Part 1 of 5 Completed: 3 of 9 My score: 3 / 9 pts (33.33%) Find the exact value of each of the remaining trigonometric functions of θ. sin θ = 12/13, θ in quadrant । 1 / 1 pt cos θ = □ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

$Andrea Myers nework #18: Section 6.3 (10 questions) Question Part 1 of 5 Completed: 3 of 9 My score: 3 / 9 pts (33.33%) Find the exact value of each of the remaining trigonometric functions of θ. sin θ = 12/13, θ in quadrant । 1 / 1 pt cos θ = □ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)$

Transcript text: Andrea Myers nework \#18: Section 6.3 (10 questions) Question Part 1 of 5 Completed: 3 of 9 My score: $3 / 9$ pts (33.33\%) Find the exact value of each of the remaining trigonometric functions of $\theta$. \[ \sin \theta=\frac{12}{13}, \theta \text { in quadrant । } \] $1 / 1 \mathrm{pt}$ \[ \boldsymbol{\operatorname { c o s }} \theta=\square \] (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Solution

Solution Steps

To find the exact value of $\cos \theta$ given $\sin \theta = \frac{12}{13}$ and $\theta$ is in the first quadrant, we can use the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. Since $\theta$ is in the first quadrant, both sine and cosine are positive. We can solve for $\cos \theta$ using this identity.

Step 1: Given Information

We are given that $\sin \theta = \frac{12}{13}$ and that $\theta$ is in the first quadrant.

Step 2: Use the Pythagorean Identity

To find $\cos \theta$, we use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting the value of $\sin \theta$: \[ \left(\frac{12}{13}\right)^2 + \cos^2 \theta = 1 \]

Step 3: Calculate $\cos^2 \theta$

Calculating $\left(\frac{12}{13}\right)^2$: \[ \frac{144}{169} \] Now substituting back into the identity: \[ \frac{144}{169} + \cos^2 \theta = 1 \] Rearranging gives: \[ \cos^2 \theta = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \]

Step 4: Calculate $\cos \theta$

Taking the square root of both sides, since $\theta$ is in the first quadrant where cosine is positive: \[ \cos \theta = \sqrt{\frac{25}{169}} = \frac{5}{13} \]