How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$. How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$?
I was thinking using slopes but I get a contradiction (i.e. $\cos(\frac{1}x) > 1$) when I do some of the algebra.
 A: use the fact that 
$$
\cos t\le 1
$$and integrate for $t$ between $0$ and $\frac 1x > 0$ to get
$$
\sin \frac 1x = \int_0^{1/x} (\cos t ) dt \le 
\int_0^{1/x} (1) dt = \frac 1x
$$
A: Let $x>1$ and consider $$\sin(y) = \sum_{n=0}^\infty \frac{(-1)^ny^{2n+1}}{(2n+1)!}$$  Substitute $y = \frac{1}{x}$: $$\sin\left(\frac{1}{x}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{x^{2n+1}(2n+1)!} = \frac{1}{x}-\frac{1}{3!x^3}+\frac{1}{5!x^5}-\frac{1}{7!x^7}+\frac{1}{9!x^9}- \dots \\ = \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots$$ now observe that each quantity in parentheses is positive, so we can think of $\sin\left(\frac{1}{x}\right)$ as $\frac{1}{x}$ minus an infinite amount of positive quantities. It follows that $$\frac{1}{x}> \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots = \sin\left(\frac{1}{x}\right)$$
A: let $f(x)=\frac{1}{x}-\sin(1/x)$ then we have $f'(x)=-\frac{1}{x^2}(1-\cos(1/x))$
A: You want to prove that $\sin \phi<\phi$ for $0<\phi<1\ (<{\pi\over2})$. The point   $P_\phi:=(\cos\phi,\sin\phi)$ on the unit circle lies in the first quadrant. Let  $A:=(1,0)$. Then $\sin\phi<|AP_\phi|$ since $\sin\phi$ is a leg of a right triangle with hypotenuse $|AP_\phi|$, and $|AP_\phi|<\phi$, since the segment $[AP_\phi]$ is shorter than the arc connecting $A$ with $P_\phi$.
A: If $x>1$, then $0<1/x<1$, so you want to prove that
$$
\sin t<t
$$
whenever $0<t<1$. Consider the function
$$
f(t)=t-\sin t
$$
Then $f(0)=0$; moreover $f'(t)=1-\cos t\ge0$ for $0<t<1$ (actually for $0<t<2\pi$), so the function $f$ is increasing in the interval $(0,1)$.
