Let $$\alpha$$ and $$\beta$$ respectively be the maximum and the minimum values of the function $$f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right),\ \ \theta\in\ R$$. Then $$\alpha+2\beta$$ is equal to:

We have, 

$$f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right)$$

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

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

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

or, $$f(\theta)=4\cos^4\theta+4\sin^4\theta-2\cos^4\theta+2\cos^2\theta\ \sin^2\theta-2\sin^4\theta$$

or, $$f(\theta)=2\cos^4\theta+2\sin^4\theta+2\cos^2\theta\ \sin^2\theta=2\left(\cos^4\theta+\sin^4\theta+\cos^2\theta\ \sin^2\theta\right)$$

or, $$f(\theta)=2\left(\cos^4\theta+\sin^4\theta+2\cos^2\theta\sin^2\theta-\cos^2\theta\sin^2\theta\right)=2\left(1-\cos^2\theta\sin^2\theta\right)$$

or, $$f(\theta)=2\left(1-\cos^2\theta\ \sin^2\theta\right)=2-2\cos^2\theta\ \sin^2\theta=2-\dfrac{\sin^22\theta}{2}\ $$

This expression has maximum and minimum values as $$\alpha$$ and $$\beta$$ respectively.

The maximum and minimum value for $$\sin^22\theta$$ is 1 and 0 respectively for $$\theta\in R\ $$. 

Hence, 

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

$$f(\theta)_{\max}=2-\dfrac{0}{2}=2\ =\alpha\ $$

So, $$\alpha+2\beta\ =2+2\times\dfrac{3}{2}=2+3=5$$