For $m>0$,
$0 < n\leqslant m+1$ ($m,n\in \mathbb{Z} $) , and $0 < a < 1$ , prove that
$$2^{n}\cdot \left( a^{n}\cos ^{2m}\dfrac {\pi a} {2}+\left( 1a\right) ^{n}\sin ^{2m}\dfrac {\pi a} {2}\right) \leqslant\cos ^{2m}\dfrac {\pi a} {2}+\sin ^{2m}\dfrac {\pi a} {2}. $$
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Note that both sides are equal at $a=1/2$, and symmetric under $a \to 1a$. So let's assume $0 < a < 1/2$. Then the inequality says $$\tan^{2m}(\pi a/2) \le \dfrac{1  2^n a^n}{2^n (1a)^n  1}$$ Now in fact, since $\tan$ is convex on $[0,\pi/4]$ we have $\tan^{2m}(\pi a/2) \le (2a)^{2m}$, and it suffices to prove that $$ 2^{2n1} a^{2n1} \le \dfrac{1  2^n a^n}{2^n (1a)^n  1}$$ i.e. that $2^{3n1}a^{2n1} (1a)^n \le 1  2^n a^n + 2^{2n1} a^{2n1}$. In fact, $a(1a) \le 1/4$, so $2^{3n1} a^{2n1}(1a)^n \le 2^{n1} a^{n1}$, and $(12^n a^n)(12^{n1}a^{n1}) \ge 0$ so $2^{n1} a^{n1} \le 1  2^n a^n + 2^{2n1} a^{2n1}$. 

