To solve for the six circular functions when $\theta = \frac{19\pi}{6}$, we need to first reduce $\theta$ to an equivalent angle between 0 and $2\pi$, and then calculate the sine, cosine, tangent, cosecant, secant, and cotangent.

Step 1: Reduce $\theta$ to an equivalent angle in the standard interval

The angle $\theta = \frac{19\pi}{6}$ is greater than $2\pi$. To find the equivalent angle within one full rotation (between 0 and $2\pi$), we subtract multiples of $2\pi$.

$\theta_{\text{equiv}} = \frac{19\pi}{6} - 2\pi = \frac{19\pi}{6} - \frac{12\pi}{6} = \frac{7\pi}{6}$

So, $\frac{19\pi}{6}$ is coterminal with $\frac{7\pi}{6}$.

Step 2: Identify the reference angle and the quadrant

The angle $\frac{7\pi}{6}$ is in the third quadrant, where sine is negative and cosine is negative. The reference angle is found by subtracting $\pi$ (which is equivalent to $6\pi/6$):

$\text{Reference angle} = \frac{7\pi}{6} - \pi = \frac{\pi}{6}$

Step 3: Calculate the six circular functions

The reference angle is $\frac{\pi}{6}$, and we know the values of sine and cosine for $\frac{\pi}{6}$ from trigonometric tables:

$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

In the third quadrant, both sine and cosine are negative. Thus:

$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}, \quad \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$

Using these, we can now calculate the other trigonometric functions.

1. Sine:

$\sin\left(\frac{19\pi}{6}\right) = \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$

2. Cosine:

$\cos\left(\frac{19\pi}{6}\right) = \cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$

3. Tangent:

$\tan\left(\frac{19\pi}{6}\right) = \frac{\sin\left(\frac{19\pi}{6}\right)}{\cos\left(\frac{19\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

4. Cosecant:

$\csc\left(\frac{19\pi}{6}\right) = \frac{1}{\sin\left(\frac{19\pi}{6}\right)} = \frac{1}{-\frac{1}{2}} = -2$

5. Secant:

$\sec\left(\frac{19\pi}{6}\right) = \frac{1}{\cos\left(\frac{19\pi}{6}\right)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$

6. Cotangent:

$\cot\left(\frac{19\pi}{6}\right) = \frac{1}{\tan\left(\frac{19\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}$

Final results:

• $\sin\left(\frac{19\pi}{6}\right) = -\frac{1}{2}$
• $\cos\left(\frac{19\pi}{6}\right) = -\frac{\sqrt{3}}{2}$
• $\tan\left(\frac{19\pi}{6}\right) = \frac{\sqrt{3}}{3}$
• $\csc\left(\frac{19\pi}{6}\right) = -2$
• $\sec\left(\frac{19\pi}{6}\right) = -\frac{2\sqrt{3}}{3}$
• $\cot\left(\frac{19\pi}{6}\right) = \sqrt{3}$

Would you like more details on any step?

Here are 5 follow-up questions for practice:

1. What are the six circular functions of $\theta = \frac{5\pi}{4}$?
2. How do you find the reference angle of any given angle in radians?
3. What is the exact value of $\tan\left(\frac{5\pi}{3}\right)$?
4. Explain why sine and cosine have different signs in the third quadrant.
5. What is the relationship between the secant and cosine functions?

Tip: Remember, for angles in the third quadrant, both sine and cosine are negative, so their reciprocals (cosecant and secant) will also be negative.