Given that $\cos(\theta) = \frac{6}{13}$ and $\sin(\theta) < 0$, we can determine the values of the trigonometric functions of $\theta$.

Step 1: Determine the quadrant

• $\cos(\theta) > 0$ (positive).
• $\sin(\theta) < 0$ (negative).

These conditions indicate that $\theta$ is in the fourth quadrant.

Step 2: Find $\sin(\theta)$

We use the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ Substitute $\cos(\theta) = \frac{6}{13}$: $\sin^2(\theta) + \left(\frac{6}{13}\right)^2 = 1$ $\sin^2(\theta) + \frac{36}{169} = 1$ $\sin^2(\theta) = 1 - \frac{36}{169}$ $\sin^2(\theta) = \frac{169}{169} - \frac{36}{169} = \frac{133}{169}$ $\sin(\theta) = \pm\sqrt{\frac{133}{169}} = \pm\frac{\sqrt{133}}{13}$

Since $\sin(\theta) < 0$ (in the fourth quadrant): $\sin(\theta) = -\frac{\sqrt{133}}{13}$

Step 3: Find $\tan(\theta)$

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{\sqrt{133}}{13}}{\frac{6}{13}} = -\frac{\sqrt{133}}{6}$

Step 4: Find $\csc(\theta)$, $\sec(\theta)$, and $\cot(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)} = -\frac{13}{\sqrt{133}}$ $\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{13}{6}$ $\cot(\theta) = \frac{1}{\tan(\theta)} = -\frac{6}{\sqrt{133}}$

Summary of Trigonometric Functions

• $\cos(\theta) = \frac{6}{13}$
• $\sin(\theta) = -\frac{\sqrt{133}}{13}$
• $\tan(\theta) = -\frac{\sqrt{133}}{6}$
• $\csc(\theta) = -\frac{13}{\sqrt{133}}$
• $\sec(\theta) = \frac{13}{6}$
• $\cot(\theta) = -\frac{6}{\sqrt{133}}$

Would you like more details or have any questions?

Related Questions

1. How can we use the unit circle to determine trigonometric function values?
2. What is the significance of the sign of sine in different quadrants?
3. How can we find the exact value of trigonometric functions for common angles?
4. Can we solve for $\theta$ given one trigonometric function value?
5. How do the Pythagorean identities relate to trigonometric functions?

Tip

Always remember the ASTC rule (All Students Take Calculus) to determine the sign of trigonometric functions in different quadrants.