How to prove that $(\sin\frac{1}{n}+\cos\frac{1}{n})\geqslant (\sin\frac{1}{n+1}+\cos\frac{1}{n+1})$

Question

How to prove, if $n>0$, that $$\left(\sin\frac{1}{n}+\cos\frac{1}{n}\right)\geqslant \left(\sin\frac{1}{n+1}+\cos\frac{1}{n+1}\right)$$

Answer 1

Let $\varphi(x)=\sin x+\cos x$ for $0<x<\pi/4$, then $\varphi'(x)=\cos x-\sin x>0$ for $0<x<\pi/4$, so for $n\geq 2$, we have $1/(n+1)<1/n$ and hence $\varphi(1/(n+1))<\varphi(1/n)$.

For \begin{align*} \sin 1+\cos 1-(\sin(1/2)+\cos(1/2))&=\sqrt{2}[\sin(1+\pi/4)-\sin(1/2+\pi/4)]\\ &=2\sqrt{2}\cos((3+\pi)/4)\sin(1/4)\\ &>0, \end{align*} note that $(3+\pi)/4<\pi/2$.

Answer 2

Hint:

1. $\displaystyle\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$;

2. $\displaystyle 0\le x_1\le x_2 \le \frac{\pi}{2} \Rightarrow 0\le \sin(x_1)\le \sin(x_2) \le 1$.

Answer 3

Assuming $n$ is positive integer, it is: $$\sin \frac{1}{n} -\sin \frac{1}{n+1}\ge \cos \frac{1}{n+1} -\cos \frac{1}{n} \iff$$ $$2\cos \frac{2n+1}{2n(n+1)} \cdot \sin \frac{1}{2n(n+1)} \ge 2\sin \frac{2n+1}{2n(n+1)}\cdot \sin \frac{1}{2n(n+1)} \iff$$ $$\cos \frac{2n+1}{2n(n+1)}\ge \sin \frac{2n+1}{2n(n+1)}, 0<\frac{2n+1}{2n(n+1)}<\frac{\pi}{4}.$$