Let’s analyze the trigonometric ratios and find their values based on the provided problem.

1. sin(60°):

   • Using the unit circle or trigonometric table, $\sin(60°) = \frac{\sqrt{3}}{2}$.

2. sin(-300°):

   • Reduce $-300^\circ$ to an equivalent positive angle by adding $360^\circ$: $-300^\circ + 360^\circ = 60^\circ$.
   • $\sin(-300^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

3. cos$\frac{\pi}{2}$:

   • The angle $\frac{\pi}{2}$ (or $90^\circ$) is on the y-axis. $\cos(90^\circ) = 0$.

4. cos$\frac{5\pi}{2}$:

   • Simplify $\frac{5\pi}{2}$ to a coterminal angle: Subtract $2\pi$ (one full revolution): $\frac{5\pi}{2} - 2\pi = \frac{\pi}{2}$.
   • Thus, $\cos\left(\frac{5\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$.

5. tan(-45°):

   • The tangent function is periodic, and $\tan(-45^\circ) = -\tan(45^\circ)$.
   • $\tan(45^\circ) = 1$, so $\tan(-45^\circ) = -1$.

6. tan(315°):

   • $315^\circ$ is in the fourth quadrant, and tangent is negative there. Subtract $360^\circ$ to get a coterminal negative angle: $315^\circ - 360^\circ = -45^\circ$.
   • Thus, $\tan(315^\circ) = \tan(-45^\circ) = -1$.

Final Values:

• sin(60°): $\frac{\sqrt{3}}{2}$
• sin(-300°): $\frac{\sqrt{3}}{2}$
• cos$\frac{\pi}{2}$: $0$
• cos$\frac{5\pi}{2}$: $0$
• tan(-45°): $-1$
• tan(315°): $-1$

Would you like me to explain any of these steps in more detail?

Related Questions:

1. What are coterminal angles, and how are they calculated?
2. How do trigonometric functions behave in different quadrants?
3. Why is the sine function periodic, and what is its period?
4. What is the relationship between radians and degrees in trigonometry?
5. How can reference angles simplify finding trig function values?

Tip:

When simplifying angles, always reduce them to a principal angle (between $0^\circ$ and $360^\circ$ or $0$ and $2\pi$). This makes it easier to compute values.