# Help with trigonometric inequality

I am trying to prove the following inequality I came across while looking at a geometric problem: $$\frac{(b-a)\sin{a}}{\sin{(b-a)}}\leq 1$$ with the constraints $$0\leq a.

I have confirmed this using software but haven't been able to prove it. Any help will be appreciated

• what's the geometric problem you came across Jan 30 '20 at 2:01

Define $$u = b-a \in [0,\frac{\pi}{2}]$$.

Rewrite the claim as $$u \sin (a) \leq \sin(u)$$. We fix $$a$$ and study $$\frac{\sin(u)}{u}$$ when $$u \in [0, \frac{\pi}{2}-a]$$.

If you differentiate $$\frac{\sin(u)}{u}$$, you will find no turning points on $$[0,\frac{\pi}{2}]$$. Thus since $$\frac{\sin(u)}{u}$$ is decreasing,

$$\frac{\sin(u)}{u} \geq \frac{\sin(\frac{\pi}{2}-a)}{\frac{\pi}{2}-a}$$.

Define $$c = \frac{\pi}{2}-a$$. We have

$$\frac{\sin(c)}{c}$$.

Our right hand side is $$\sin(a) = \cos(c)$$.

Thus the claim follows since $$\sin(c) \geq c \cos(c)$$ on $$[0,\frac{\pi}{2}]$$.

• Nice approach, thank you
– user92596
Jan 30 '20 at 2:48
• Your post seems unclear from when you introduce $c$ and onward. Could you elaborate? Jan 30 '20 at 3:28
• note that $\sin(c) = \cos(a)$, $\cos(c) = \sin(a)$. We want to prove that $\frac{\sin(\frac{\pi}{2}-a)}{\frac{\pi}{2}-a} \geq \sin(a)$. This is equivalent to proving $\frac{\sin(c)}{c} \geq \sin(a) = \cos(c)$. Jan 30 '20 at 3:32
• I understand all of that. What let's us say the last inequality? Jan 30 '20 at 3:40
• It's a standard one. You can prove it by differentiating $\sin(c)-c\cos(c)$ and proving it is positive on this interval. Notice that the inequality holds at $c=0$. Jan 30 '20 at 4:26

I hope this helps. On $$[0,\pi/2]$$ the function $$t\mapsto \frac{\sin t}{t}$$ is decreasing, and $$t\mapsto \sin t$$ is increasing. So, if $$0\leq a,

$$\frac{\sin a}{b}\leq \frac{\sin b}{b}\leq \frac{\sin(b-a)}{b-a}$$

Hence $$\frac{(b-a)\sin a}{\sin(b-a)}\leq b$$

The inequality you suggested follows for $$b\leq 1$$. For the other cases it still escapes me.

• Nice idea, thanks for the help.
– user92596
Jan 30 '20 at 2:49