Questions: Find the exact value of cos θ. tan θ=-10/7, θ in quadrant II A. sqrt(149)/10 B. 7 sqrt(149)/149 C. -sqrt(149)/7 D. -7 sqrt(149)/149

Transcript text: Find the exact value of $\cos \theta$. $\tan \theta=-\frac{10}{7}, \theta$ in quadrant II A. $\frac{\sqrt{149}}{10}$ B. $\frac{7 \sqrt{149}}{149}$ C. $-\frac{\sqrt{149}}{7}$ D. $-\frac{7 \sqrt{149}}{149}$

Solution

Solution Steps

To find the exact value of $\cos \theta$ given $\tan \theta = -\frac{10}{7}$ and $\theta$ is in quadrant II, we can use the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$. In quadrant II, $\sin \theta$ is positive and $\cos \theta$ is negative. We can use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$.

Express $\sin \theta$ in terms of $\cos \theta$ using $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
Substitute $\sin \theta$ in the Pythagorean identity to solve for $\cos \theta$.
Determine the sign of $\cos \theta$ based on the quadrant.

Step 1: Express $\sin \theta$ in terms of $\cos \theta$

Given that $\tan \theta = -\frac{10}{7}$, we can express $\sin \theta$ as: \[ \sin \theta = \tan \theta \cdot \cos \theta = -\frac{10}{7} \cdot \cos \theta \]

Step 2: Use the Pythagorean Identity

Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we substitute $\sin \theta$: \[ \left(-\frac{10}{7} \cdot \cos \theta\right)^2 + \cos^2 \theta = 1 \] This simplifies to: \[ \frac{100}{49} \cdot \cos^2 \theta + \cos^2 \theta = 1 \] Combining the terms gives: \[ \left(\frac{100}{49} + 1\right) \cos^2 \theta = 1 \] \[ \frac{149}{49} \cos^2 \theta = 1 \]

Step 3: Solve for $\cos \theta$

Rearranging the equation, we find: \[ \cos^2 \theta = \frac{49}{149} \] Taking the square root, we have: \[ \cos \theta = \pm \sqrt{\frac{49}{149}} = \pm \frac{7}{\sqrt{149}} \] Since $\theta$ is in quadrant II, where $\cos \theta$ is negative, we conclude: \[ \cos \theta = -\frac{7}{\sqrt{149}} \]