High school inequality |(√(sinx)+1)^2−(√(sina)+1)^2|<b

[solakis1]

Given 0<a<π/2 , b>0 find a c>0 such that :

for all ,x : if 0<x<π/2 and |x-a|<c ,then $$|(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$

 

[Opalg]

Given $0<a<\pi/2$, $b>0$, find a $c>0$ such that :

for all $x$ : if $0<x<\pi/2$ and $|x-a|<c $, then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2|<b$$.

[sp]First step: $|\sin x - \sin a| < |x-a|$. That is essentially the Mean Value theorem, which says that $\sin x - \sin a = (x-a)\cos y$ for some $y$ between $a$ and $x$.

Next, $|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| = |\sin x + 2\sqrt{\sin x} + 1 - (\sin a + 2\sqrt{\sin a} + 1)| = |(\sin x -\sin a) + 2(\sqrt{\sin x} - \sqrt{\sin a})| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}|.$

To get an estimate for $\sqrt{\sin x} - \sqrt{\sin a}$, multiply top and bottom by $\sqrt{\sin x} + \sqrt{\sin a}$: $$\sqrt{\sin x} - \sqrt{\sin a} = \frac{(\sqrt{\sin x} - \sqrt{\sin a})(\sqrt{\sin x} + \sqrt{\sin a})}{\sqrt{\sin x} + \sqrt{\sin a}} = \frac{\sin x - \sin a}{\sqrt{\sin x} + \sqrt{\sin a}},$$ from which $$|\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

Then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).$$

We want this to be less than $b$. So take $$c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
[/sp]

 

[Jameson]

Given 0<a<π/2 , b>0 find a c>0 such that :

for all ,x : if 0<x<π/2 and |x-a|<c ,then $$|(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$

Hi solakis,

Is this a question that you want help with or know the answer to already and want to challenge others? Either way is fine but if it's the former then I'll move the thread to a different part of the forum.

 

[solakis1]

Hi solakis,

Is this a question that you want help with or know the answer to already and want to challenge others? Either way is fine but if it's the former then I'll move the thread to a different part of the forum.

This is a challenge question

 

[solakis1]

[sp]First step: $|\sin x - \sin a| < |x-a|$. That is essentially the Mean Value theorem, which says that $\sin x - \sin a = (x-a)\cos y$ for some $y$ between $a$ and $x$.

Next, $|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| = |\sin x + 2\sqrt{\sin x} + 1 - (\sin a + 2\sqrt{\sin a} + 1)| = |(\sin x -\sin a) + 2(\sqrt{\sin x} - \sqrt{\sin a})| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}|.$

To get an estimate for $\sqrt{\sin x} - \sqrt{\sin a}$, multiply top and bottom by $\sqrt{\sin x} + \sqrt{\sin a}$: $$\sqrt{\sin x} - \sqrt{\sin a} = \frac{(\sqrt{\sin x} - \sqrt{\sin a})(\sqrt{\sin x} + \sqrt{\sin a})}{\sqrt{\sin x} + \sqrt{\sin a}} = \frac{\sin x - \sin a}{\sqrt{\sin x} + \sqrt{\sin a}},$$ from which $$|\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

Then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).$$

We want this to be less than $b$. So take $$c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
[/sp]

[sp] This supose to be a high school inequality :

|sinx-sina|=$$2|cos\frac{x+a}{2}|.|sin\frac{x-a}{2}|
\leq 2|sin\frac{x-a}{2}|$$....because $$|cost|\leq 1\forall t$$$$\leq 2|\frac{x-a}{2}|=|x-a|$$...since $$|sint|\leq |t|\forall t$$.But i am wondering do they learn the inequality $$|sint|\leq | t|\forall t $$ at high school ?[/sp]