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

• MHB
• solakis1
In summary, we are trying to find a value for c that satisfies the inequality for all x, given the values of a and b. Using the Mean Value theorem, we can simplify the expression and get an estimate for the value of c. We then use the fact that |sint| ≤ |t| for all t to find a more precise value for c. The use of this inequality may be taught in high school.
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 $$\displaystyle |(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$

solakis said:
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 $$\displaystyle |(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2|<b$$.
[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 $$\displaystyle |\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

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

We want this to be less than $b$. So take $$\displaystyle c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
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solakis said:
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 $$\displaystyle |(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$

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.

Jameson said:
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

Opalg said:
[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 $$\displaystyle |\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

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

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

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

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