Trigonometric Function Simplification

First, simplify the trigonometric terms in the function $f(\theta)$ using angle properties and periodicity:

• $\sin\left(\frac{7\pi}{2} - \theta\right) = \sin\left(3\pi + \frac{\pi}{2} - \theta\right) = \sin\left(\pi + \left(\frac{\pi}{2} - \theta\right)\right) = -\sin\left(\frac{\pi}{2} - \theta\right) = -\cos(\theta)$
• $\sin(11\pi + \theta) = \sin(\pi + \theta) = -\sin(\theta)$
• $\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos(\theta)$
• $\sin(9\pi - \theta) = \sin(\pi - \theta) = \sin(\theta)$

Substitute these back into the function:

$f(\theta) = 4\left((-\cos\theta)^4 + (-\sin\theta)^4\right) - 2\left((-\cos\theta)^6 + (\sin\theta)^6\right)$

$f(\theta) = 4\left(\cos^4\theta + \sin^4\theta\right) - 2\left(\cos^6\theta + \sin^6\theta\right)$

Use the identities:

• $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta$
• $\cos^6\theta + \sin^6\theta = 1 - 3\sin^2\theta\cos^2\theta$

Substitute these identities:

$f(\theta) = 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$

$f(\theta) = 4 - 8\sin^2\theta\cos^2\theta - 2 + 6\sin^2\theta\cos^2\theta$

$f(\theta) = 2 - 2\sin^2\theta\cos^2\theta$

Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we get $\sin^2\theta\cos^2\theta = \frac{\sin^2(2\theta)}{4}$:

$f(\theta) = 2 - 2\left(\frac{\sin^2(2\theta)}{4}\right) = 2 - \frac{1}{2}\sin^2(2\theta)$

Finding Extrema $(\alpha, \beta)$

The simplified function is $f(\theta) = 2 - \frac{1}{2}\sin^2(2\theta)$.

We know that the range of $\sin(x)$ is $[-1, 1]$, so the range of $\sin^2(x)$ is $[0, 1]$.

Therefore, the range of $\sin^2(2\theta)$ is $[0, 1]$.

To find the maximum value ($\alpha$), we minimize $\sin^2(2\theta)$:

$\alpha = \max(f(\theta)) = 2 - \frac{1}{2}(0) = 2$

To find the minimum value ($\beta$), we maximize $\sin^2(2\theta)$:

$\beta = \min(f(\theta)) = 2 - \frac{1}{2}(1) = 2 - \frac{1}{2} = \frac{3}{2}$

Final Calculation

The question asks for the value of $\alpha + 2\beta$.

Substitute the found values of $\alpha$ and $\beta$:

$\alpha + 2\beta = 2 + 2\left(\frac{3}{2}\right)$

$\alpha + 2\beta = 2 + 3 = 5$