If $\sin (\theta)=-\frac{5}{6}$ And $\theta$ Is In The 3rd Quadrant, Find $\cos (\theta$\].$\cos (\theta) = \square$Enter Your Answer As A Reduced Radical. Enter $\sqrt{12}$ As 2 Sqrt(3).

If $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant, find $\cos (\theta)$

In trigonometry, the sine and cosine functions are used to describe the relationships between the angles and side lengths of triangles. Given the value of the sine function, we can use the Pythagorean identity to find the value of the cosine function. In this article, we will explore how to find the value of $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant.

The Pythagorean identity states that for any angle $\theta$, the following equation holds:

$$\sin^2 (\theta) + \cos^2 (\theta) = 1$$

This identity can be used to find the value of $\cos (\theta)$ when the value of $\sin (\theta)$ is known.

$$

Given that $\sin (\theta)=-\frac{5}{6}$, we can substitute this value into the Pythagorean identity:

$$\left(-\frac{5}{6}\right)^2 + \cos^2 (\theta) = 1$$

Simplifying the left-hand side of the equation, we get:

$$\frac{25}{36} + \cos^2 (\theta) = 1$$

Subtracting $\frac{25}{36}$ from both sides of the equation, we get:

$$\cos^2 (\theta) = 1 - \frac{25}{36}$$

Simplifying the right-hand side of the equation, we get:

$$\cos^2 (\theta) = \frac{11}{36}$$

Taking the square root of both sides of the equation, we get:

$$\cos (\theta) = \pm \sqrt{\frac{11}{36}}$$

Since $\theta$ is in the 3rd quadrant, the cosine function is negative. Therefore, we can write:

$$\cos (\theta) = -\sqrt{\frac{11}{36}}$$

To simplify the answer, we can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{36}$:

$$\cos (\theta) = -\frac{\sqrt{11}}{\sqrt{36}} \cdot \frac{\sqrt{36}}{\sqrt{36}}$$

Simplifying the expression, we get:

$$\cos (\theta) = -\frac{\sqrt{11 \cdot 36}}{36}$$

Simplifying the numerator, we get:

$$\cos (\theta) = -\frac{\sqrt{396}}{36}$$

Simplifying the square root, we get:

$$\cos (\theta) = -\frac{\sqrt{4 \cdot 99}}{36}$$

Simplifying the expression, we get:

$$\cos (\theta) = -\frac{2\sqrt{99}}{36}$$

Simplifying the square root, we get:

$$\cos (\theta) = -\frac{2\sqrt{9 \cdot 11}}{36}$$

Simplifying the expression, we get:

$$\cos (\theta) = -\frac{2 \cdot 3\sqrt{11}}{36}$$

Simplifying the expression, we get:

$$\cos (\theta) = -\frac{6\sqrt{11}}{36}$$

Simplifying the fraction, we get:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

In this article, we used the Pythagorean identity to find the value of $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant. We simplified the answer to get:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

This is the final answer.
Q&A: Finding $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant

In our previous article, we explored how to find the value of $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant. We used the Pythagorean identity to find the value of $\cos (\theta)$ and simplified the answer to get:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

In this article, we will answer some common questions related to finding $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant.

Q: What is the Pythagorean identity?

A: The Pythagorean identity is a fundamental concept in trigonometry that states:

$$\sin^2 (\theta) + \cos^2 (\theta) = 1$$

This identity can be used to find the value of $\cos (\theta)$ when the value of $\sin (\theta)$ is known.

Q: How do I find $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant?

A: To find $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant, you can use the Pythagorean identity:

$$\left(-\frac{5}{6}\right)^2 + \cos^2 (\theta) = 1$$

Simplifying the left-hand side of the equation, you get:

$$\frac{25}{36} + \cos^2 (\theta) = 1$$

Subtracting $\frac{25}{36}$ from both sides of the equation, you get:

$$\cos^2 (\theta) = 1 - \frac{25}{36}$$

Simplifying the right-hand side of the equation, you get:

$$\cos^2 (\theta) = \frac{11}{36}$$

Taking the square root of both sides of the equation, you get:

$$\cos (\theta) = \pm \sqrt{\frac{11}{36}}$$

Since $\theta$ is in the 3rd quadrant, the cosine function is negative. Therefore, you can write:

$$\cos (\theta) = -\sqrt{\frac{11}{36}}$$

Q: How do I simplify the answer?

A: To simplify the answer, you can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{36}$:

$$\cos (\theta) = -\frac{\sqrt{11}}{\sqrt{36}} \cdot \frac{\sqrt{36}}{\sqrt{36}}$$

Simplifying the expression, you get:

$$\cos (\theta) = -\frac{\sqrt{11 \cdot 36}}{36}$$

Simplifying the numerator, you get:

$$\cos (\theta) = -\frac{\sqrt{396}}{36}$$

Simplifying the square root, you get:

$$\cos (\theta) = -\frac{\sqrt{4 \cdot 99}}{36}$$

Simplifying the expression, you get:

$$\cos (\theta) = -\frac{2\sqrt{99}}{36}$$

Simplifying the square root, you get:

$$\cos (\theta) = -\frac{2\sqrt{9 \cdot 11}}{36}$$

Simplifying the expression, you get:

$$\cos (\theta) = -\frac{2 \cdot 3\sqrt{11}}{36}$$

Simplifying the expression, you get:

$$\cos (\theta) = -\frac{6\sqrt{11}}{36}$$

Simplifying the fraction, you get:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

Q: What is the final answer?

A: The final answer is:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

In this article, we answered some common questions related to finding $\cos (\theta)$ when $\sin (\theta)=-\frac{5}{6}$ and $\theta$ is in the 3rd quadrant. We used the Pythagorean identity to find the value of $\cos (\theta)$ and simplified the answer to get:

$$\cos (\theta) = -\frac{\sqrt{11}}{6}$$

This is the final answer.