The problem in your image asks to find the exact values of the six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) for the angle $420^\circ$.

Step-by-Step Solution:

1. Standardize the angle: To work with angles, we first reduce $420^\circ$ to an equivalent angle within the standard $0^\circ$ to $360^\circ$ range by subtracting $360^\circ$: $420^\circ - 360^\circ = 60^\circ$ So, $\sin(420^\circ) = \sin(60^\circ)$, and similarly for the other functions.

2. Use known trigonometric values for $60^\circ$: From the unit circle, the trigonometric values for $60^\circ$ are well-known: $\sin(60^\circ) = \frac{\sqrt{3}}{2}, \quad \cos(60^\circ) = \frac{1}{2}, \quad \tan(60^\circ) = \sqrt{3}$ Now we can find the remaining functions: $\csc(60^\circ) = \frac{1}{\sin(60^\circ)} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}, \quad \sec(60^\circ) = \frac{1}{\cos(60^\circ)} = 2$ $\cot(60^\circ) = \frac{1}{\tan(60^\circ)} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

3. Conclusion: The exact values of the six trigonometric functions for $420^\circ$ (or $60^\circ$) are:

   • $\sin(420^\circ) = \frac{\sqrt{3}}{2}$
   • $\cos(420^\circ) = \frac{1}{2}$
   • $\tan(420^\circ) = \sqrt{3}$
   • $\csc(420^\circ) = \frac{2\sqrt{3}}{3}$
   • $\sec(420^\circ) = 2$
   • $\cot(420^\circ) = \frac{\sqrt{3}}{3}$

Would you like further explanation on any step or the application of trigonometric identities?

Relative Questions:

1. How do we calculate trigonometric functions for angles greater than $360^\circ$?
2. Why do we subtract $360^\circ$ to reduce the angle to a standard form?
3. What is the importance of the unit circle in trigonometry?
4. How do you find the trigonometric values for angles like $45^\circ$ or $30^\circ$?
5. What is the relationship between the sine and cosecant functions?

Tip:

Always simplify angles greater than $360^\circ$ by subtracting multiples of $360^\circ$ to bring the angle within the first full rotation of the unit circle!