It is easy to prove that $\sin(a) < \tan(a)$ when $0 < a < \pi/2$, but how can I prove that $\sin(a) < a < \tan(a)$ when $0 < a < \pi/2?$
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this is slick – AgentS Aug 08 '19 at 07:38
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1In geometry, $\sin(a)$ is the perpendicular straight line distance from a point on the unit circle to a radius of the circle and is therefore shorter than $a$ which is a curved distance from the same point to the same radius – Henry Aug 08 '19 at 07:38
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1This is essentially inequality $(1)$ from this answer. – robjohn Aug 08 '19 at 07:53
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For $0 \le a \lt \frac{\pi}{2}$, define
$$f(a) = a - \sin(a) \tag{1}\label{eq1}$$
$$g(a) = \tan(a) - a \tag{2}\label{eq2}$$
From \eqref{eq1}, note $f(0) = 0$. For $a \gt 0$, $f'(a) = 1 - \cos(a) \gt 0$ so $f(a) \gt 0$, giving
$$\sin(a) \lt a \tag{3}\label{eq3}$$
From \eqref{eq2}, $g(0) = 0$. For $a \gt 0$, $g'(a) = \frac{\cos(a)}{\cos(a)} + \frac{\sin^2(a)}{\cos^2(a)} - 1 = \frac{\sin^2(a)}{\cos^2(a)} \gt 0$ so $g(a) \gt 0$, giving
$$a \lt \tan(a) \tag{4}\label{eq4}$$
Putting \eqref{eq3} and \eqref{eq4} together gives
$$\sin(a) \lt a \lt \tan(a) \tag{5}\label{eq5}$$
for $0 \lt a \lt \frac{\pi}{2}$.
John Omielan
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@Kevin I'm glad to know the original was good enough for you. I just thought adding a bit more detail would make it somewhat clearer and easier to understand. – John Omielan Aug 08 '19 at 07:31
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Let $$f(x)=\sin x-x \rightarrow f'(x)=\cos x-1 \le 0.$$ This means f$(x)$ is a decreasing function in $[0,\pi/2]$, so $f(x) \le f(0) \Rightarrow \sin x \le x, x \in [0,\pi/2].$$
Next take $$g(x)=\tan x -x \Rightarrow g'(x)=\sec^2 x-1\ge 0.$$ So $g(x)$ is an increasing fumction for $x \in [0, \infty).$ Then $$g(x) \ge g(0) \Rightarrow \tan x \ge x, x\in[0,\infty). $$
Z Ahmed
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