How can I prove that $\frac{\sin\theta+ \tan\theta}{2}> \theta$ when $0 < \theta < \frac{\pi}{2}$

Question

I don’t know how to do it without calculus, could someone help ? How do I prove it geometrically and/or algebraically ?

EDIT: $0 < \theta < \frac{\pi}{2}$

My teacher never explained it. Someone tried explaining it to me with calculus, but I would appreciate a different explanation.

Do you have bounds on $\theta$, perhaps a specific interval to check? Otherwise this inequality is false infinitely often... — abiessu, Mar 09 '21 at 06:19
Please state in your post what kind of methods you have tried so we know where you got stuck. — Lord Commander, Mar 09 '21 at 06:19
Here's a graph of it if anyone wants: https://www.geogebra.org/classic/premqyks — Mike Smith, Mar 09 '21 at 06:33
Please see how to ask a good question. This has a list of ways you can add context (supporting information) to your questions. — Toby Mak, Mar 09 '21 at 06:53
I’m not sure if there is a geometric way but thank you for your help. I will look at how to ask a good question for the future. — sandwichlover42, Mar 09 '21 at 07:12
There is a geometric way: click the link called "Is there a geometrical method...". If you use Approach0, you will often find the answer to your question already there. — Toby Mak, Mar 09 '21 at 07:14

score 3 · Accepted Answer · answered Mar 09 '21 at 06:32

3

Since $\sin \theta, \tan \theta > 0$ in the given domain, AM > HM so we have:

$$\frac{\sin\theta+ \tan\theta}{2} > \frac{2}{1/\sin \theta + 1/\tan \theta}$$

and $\csc \theta + \cot \theta = \cot(\theta/2)$, thus the RHS is $2 \tan(\theta/2) > 2 \cdot \theta/2 = \theta$.

In the last line, we have used the fact that $\tan x > x$ when $0 < x < \pi/2$.

answered Mar 09 '21 at 06:32

Toby Mak

1

You can get the identity for $\csc \theta + \cot \theta$ by finding $\cot^2(\theta/2) = \cos^2(\theta/2) / \sin^2(\theta/2) = \frac{1 + \cos \theta}{1 - \cos \theta}$. Now divide top and bottom by $\sin \theta$ and multiply by the conjugate. Using the result $\csc^2 x - \cot^2 x = 1$ gives the result. – Toby Mak Mar 09 '21 at 06:40
thank you so much! This really helped – sandwichlover42 Mar 09 '21 at 07:11

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