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(B) Given $f(	heta)=4(\sin^{4}(\frac{7\pi}{2}-	heta)+\sin^{4}(11\pi+	heta)) - 2(\sin^{6}(\frac{3\pi}{2}-	heta)+\sin^{6}(9\pi-	heta))$.Using trigo

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$5$

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(B) Given $f(	heta)=4(\sin^{4}(\frac{7\pi}{2}-	heta)+\sin^{4}(11\pi+	heta)) - 2(\sin^{6}(\frac{3\pi}{2}-	heta)+\sin^{6}(9\pi-	heta))$.Using trigonometric identities:$\sin(\frac{7\pi}{2}-	heta) = \cos	heta$ and $\sin(11\pi+	heta) = -\sin	heta$.$\sin(\frac{3\pi}{2}-	heta) = -\cos	heta$ and $\sin(9\pi-	heta) = \sin	heta$.Substituting these, we get:$f(	heta) = 4(\cos^{4}	heta + \sin^{4}	heta) - 2(\cos^{6}	heta + \sin^{6}	heta)$.Using the identities $\sin^{4}	heta + \cos^{4}	heta = 1 - 2\sin^{2}	heta\cos^{2}	heta$ and $\sin^{6}	heta + \cos^{6}	heta = 1 - 3\sin^{2}	heta\cos^{2}	heta$:$f(	heta) = 4(1 - 2\sin^{2}	heta\cos^{2}	heta) - 2(1 - 3\sin^{2}	heta\cos^{2}	heta) = 4 - 8\sin^{2}	heta\cos^{2}	heta - 2 + 6\sin^{2}	heta\cos^{2}	heta = 2 - 2\sin^{2}	heta\cos^{2}	heta$.Since $\sin^{2}	heta\cos^{2}	heta = \frac{(2\sin	heta\cos	heta)^{2}}{4} = \frac{\sin^{2}(2	heta)}{4}$, we have:$f(	heta) = 2 - 2(\frac{\sin^{2}(2	heta)}{4}) = 2 - \frac{\sin^{2}(2	heta)}{2}$.The maximum value $\alpha$ occurs when $\sin^{2}(2	heta) = 0$, so $\alpha = 2$.The minimum value $\beta$ occurs when $\sin^{2}(2	heta) = 1$, so $\beta = 2 - \frac{1}{2} = \frac{3}{2}$.Therefore, $\alpha + 2\beta = 2 + 2(\frac{3}{2}) = 2 + 3 = 5$.

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

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