Question

The minimum value of \[ \left(1+\frac{1}{\sin^n\alpha}\right) \left(1+\frac{1}{\cos^n\alpha}\right) \] is

• A. \(1\)
• B. \(2\)
• C. \(\left(1+2^n\right)^2\)
• D. \(\left(1+2^{\frac{n}{2}}\right)^2\)

Hint:

When an expression is symmetric in \(\sin\alpha\) and \(\cos\alpha\), its minimum or maximum generally occurs at \(\sin\alpha=\cos\alpha\), provided the expression is defined.

Answer 1

The Correct Option is D

Step 1: Expand the product.
Let \(E=\left(1+\frac{1}{\sin^n\alpha}\right)\left(1+\frac{1}{\cos^n\alpha}\right) = 1+\frac{1}{\sin^n\alpha}+\frac{1}{\cos^n\alpha}+\frac{1}{(\sin\alpha\cos\alpha)^n}\)

Step 2: Bound the middle two terms using AM-GM.
By AM-GM, \(\frac{1}{\sin^n\alpha}+\frac{1}{\cos^n\alpha}\geq\frac{2}{(\sin\alpha\cos\alpha)^{n/2}}\), with equality exactly when \(\sin\alpha=\cos\alpha\).

Step 3: Bound the last term.
Since \(\sin\alpha\cos\alpha=\frac{1}{2}\sin2\alpha\leq\frac{1}{2}\), the term \(\frac{1}{(\sin\alpha\cos\alpha)^n}\) is smallest exactly when \(\sin\alpha\cos\alpha\) is largest, again at \(\sin\alpha=\cos\alpha=\frac{1}{\sqrt2}\) (so \(\sin\alpha\cos\alpha=\frac12\)).

Step 4: Both bounds hit their best case together.
At \(\alpha=\pi/4\), \(\sin\alpha\cos\alpha=\frac12\), so \(\frac{1}{\sin^n\alpha}+\frac{1}{\cos^n\alpha}=2\cdot2^{n/2}\) and \(\frac{1}{(\sin\alpha\cos\alpha)^n}=2^n\).
\[ E_{min}=1+2\cdot2^{n/2}+2^n=\left(1+2^{n/2}\right)^2 \]

Step 5: Final conclusion.
\[ \boxed{\left(1+2^{n/2}\right)^2} \]