The image you provided contains a graph of the trigonometric function and a question asking for the value of $C$ in the equation $C \cdot \cos(5\alpha)$. Let's break down the information and solve the problem.

Key information from the image:

1. The given function is in the form: $y = -2 \cdot \cos(k(x - \alpha)) + d$ with constraints for $\alpha$ in degrees: $-90^\circ < \alpha < 90^\circ$.

2. The graph is periodic and shows a cosine wave, with notable points at $x = 60^\circ$ and $x = 210^\circ$, implying a half-cycle from $60^\circ$ to $210^\circ$.

3. The graph shows that the maximum value is $y = 5$ and the minimum value is $y = 1$, which gives us clues about the vertical shift ($d$).

Solving the problem:

From the form of the equation $y = -2 \cdot \cos(k(x - \alpha)) + d$:

• Amplitude: The amplitude of the function is the coefficient in front of the cosine function, which is $|-2| = 2$. The difference between the maximum and minimum values on the graph is $5 - 1 = 4$, confirming that the amplitude is 2.

• Vertical Shift: The midline of the graph can be found as the average of the maximum and minimum values, which is: $\text{midline} = \frac{5 + 1}{2} = 3$ So, the vertical shift $d = 3$.

• Period: The period of a cosine function is given by $\frac{360^\circ}{k}$. From the graph, it takes $210^\circ - 60^\circ = 150^\circ$ for half a cycle, so a full cycle would be $2 \times 150^\circ = 300^\circ$. Thus: $\frac{360^\circ}{k} = 300^\circ \implies k = \frac{360^\circ}{300^\circ} = 1.2$

• Phase Shift: The graph crosses the midline at $x = 60^\circ$, which is a clue for the phase shift $\alpha$. Since cosine typically starts at a maximum at $x = 0^\circ$, the phase shift $\alpha = 60^\circ$.

Thus, the equation of the graph becomes: $y = -2 \cdot \cos(1.2(x - 60^\circ)) + 3$

Now, to find the value of $C \cdot \cos(5\alpha)$:

• Given $\alpha = 60^\circ$, we need to calculate $C \cdot \cos(5 \cdot 60^\circ) = C \cdot \cos(300^\circ)$.
• $\cos(300^\circ) = \frac{1}{2}$, so: $C \cdot \frac{1}{2} = 1 \implies C = 2$

Thus, the value of $C = 2$.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How do you determine the amplitude of a trigonometric function from its graph?
2. What is the significance of the vertical shift in trigonometric graphs?
3. How does the phase shift affect the starting point of a cosine graph?
4. What is the period of a cosine function and how is it calculated from the graph?
5. How can trigonometric identities simplify solving problems like this?

Tip: Always check the key points on the graph (maximum, minimum, midline) to derive the amplitude, period, and phase shift easily.