Given that the point $P = \left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right)$ corresponds to the real number $t$ on the unit circle, we can use the coordinates of this point to find the values of the six trigonometric functions. Recall that the unit circle gives us the relationships:

• $\sin t = y$ (the $y$-coordinate)
• $\cos t = x$ (the $x$-coordinate)
• $\tan t = \frac{y}{x}$
• $\csc t = \frac{1}{y}$
• $\sec t = \frac{1}{x}$
• $\cot t = \frac{x}{y}$

Step-by-step Calculation:

1. Sine (sin t): $\sin t = y = \frac{\sqrt{3}}{2}$

2. Cosine (cos t): $\cos t = x = -\frac{1}{2}$

3. Tangent (tan t): $\tan t = \frac{y}{x} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$

4. Cosecant (csc t): $\csc t = \frac{1}{y} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

5. Secant (sec t): $\sec t = \frac{1}{x} = \frac{1}{-\frac{1}{2}} = -2$

6. Cotangent (cot t): $\cot t = \frac{x}{y} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

Final Values:

• $\sin t = \frac{\sqrt{3}}{2}$
• $\cos t = -\frac{1}{2}$
• $\tan t = -\sqrt{3}$
• $\csc t = \frac{2\sqrt{3}}{3}$
• $\sec t = -2$
• $\cot t = -\frac{\sqrt{3}}{3}$

Would you like more explanation on any of these calculations, or do you have any further questions?

Here are five related questions:

1. What is the general form of the unit circle equations for trigonometric functions?
2. How do we determine the signs of the trigonometric functions in different quadrants of the unit circle?
3. How do you convert between the six trigonometric functions?
4. What are the Pythagorean identities for trigonometric functions?
5. How can we use the unit circle to solve for unknown angles in trigonometric equations?

Tip: Remember that in the unit circle, the sine function is the $y$-coordinate and the cosine function is the $x$-coordinate.