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Q. Statement-1: Minimum value of function $f(\theta)=\sin ^4 \theta+\frac{\sec \theta}{8 \tan \theta}$ is $\frac{5}{16}$ for $\frac{\pi}{2}<\theta<\pi$. Let $a _{ i }>0 \quad \forall i \in N$, then $\frac{ a _1+ a _2+ a _3+\ldots \ldots . .+ a _{ n }}{ n } \geq \sqrt[n]{ a _1 a _2 a _3 \ldots \ldots a _{ n }}$
Statement-2: where equality is obtained at $a _1= a _2= a _3 \ldots \ldots \ldots= a _{ n }$.

Sequences and Series

Solution:

Given $f(\theta)=\sin ^4 \theta+\frac{\sec \theta}{8 \tan \theta}=\sin ^4 \theta+\frac{\operatorname{cosec} \theta}{8}=\sin ^4 \theta+\frac{1}{8 \sin \theta}$ As $\sin ^4 \theta, \frac{1}{32 \sin \theta}, \frac{1}{32 \sin \theta}, \frac{1}{32 \sin \theta}, \frac{1}{32 \sin \theta}$ are positive in $\left(\frac{\pi}{2}, \pi\right)$
Now using A.M. $\geq$ G.M., we get
$\frac{\sin ^4 \theta+\frac{1}{8 \sin \theta}}{5} \geq\left(\frac{1}{(32)^4}\right)^{1 / 5} \text { or } \sin ^4 \theta+\frac{1}{8} \operatorname{cosec} \theta \geq \frac{5}{16}$
Hence minimum value of $\sin ^4 \theta+\frac{1}{8} \operatorname{cosec} \theta$ is $\frac{5}{16}$