That's So Min!

Let $sin \alpha = x$ and $cos \alpha = y$ , satisfying $x^2+y^2 = 1$ . We are then looking to minimize $x^{2016}+y^{2016}$ .

By Holder's inequality,

$\large ((1+1)(1+1) \cdots (1+1)(x^{2016}+y^{2016}))^{\frac{1}{1008}} \geq x^\frac{2016}{1008}+y^\frac{2016}{1008}$

With $1007$ $1+1$ 's in the LHS.

Simplifying gives us $2^{1007}(x^{2016}+y^{2016}) \geq (x^2+y^2)^{1008} = 1$ (since $x^2+y^2 = 1$ )

Giving us $x^{2016}+y^{2016} \geq \frac{1}{2^{1007}} = 2^{-1007}$ .

So, our minimum value of $sin^{2016}\alpha + cos^{2016}\alpha = x^{2016}+y^{2016}$ is $2^{-1007}$ . We then get our answer as $\boxed{-1007}$ .

First, note that $sin^{2016} \alpha$ and $cos^{2016} \alpha$ are non-negative. For cases where either $sin \alpha=0$ or $cos \alpha=0$ , we have $sin^{2016} \alpha + cos^{2016} \alpha =1$ . For positive values, we apply AM-GM inequality(?):

$sin^{2016} \alpha + cos^{2016} \alpha \geq 2 \cdot sin^{1008} \alpha \cdot cos^{1008} \alpha$

We take the equality case for minimum value. This gives us $sin\alpha=cos\alpha=\frac{1}{\sqrt{2}}$ . By substitution, we have $sin^{2016} \alpha + cos^{2016} \alpha = \frac{1}{2^{1007}}$ . Hence $\boxed{n=-1007}$ .