Prove $sin^{-1}(1)geq int_0^b1/sqrt{1-x^2}dx +(1-b)pi/2$ for $b in [0,1)$
$begingroup$
I'm trying to prove the following inequality:
$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$
for every $b in [0,1)$.
I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).
My Attempt
I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.
This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.
I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.
real-analysis integration trigonometry inequality proof-writing
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following inequality:
$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$
for every $b in [0,1)$.
I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).
My Attempt
I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.
This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.
I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.
real-analysis integration trigonometry inequality proof-writing
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
$endgroup$
3
$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07
add a comment |
$begingroup$
I'm trying to prove the following inequality:
$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$
for every $b in [0,1)$.
I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).
My Attempt
I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.
This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.
I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.
real-analysis integration trigonometry inequality proof-writing
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
$endgroup$
I'm trying to prove the following inequality:
$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$
for every $b in [0,1)$.
I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).
My Attempt
I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.
This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.
I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.
real-analysis integration trigonometry inequality proof-writing
real-analysis integration trigonometry inequality proof-writing
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
edited Dec 12 '18 at 7:44
Sameer Baheti
5168
5168
asked Dec 11 '18 at 17:39
user624612
3
$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07
add a comment |
3
$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07
3
3
$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
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add a comment |
$begingroup$
I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
The inequality you seek to prove is not true. Put $b=1/4$.
However, $sin^{-1}ble bpi/2, forall bin[0,1)$
Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$
$f'(x)=pi/2-frac1{sqrt{1-x^2}}$
$f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$
This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
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add a comment |
$begingroup$
If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$ This implies
$$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$
for $bin (0,1).$ Apply this with $f(b)=arcsin b.$
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
$endgroup$
add a comment |
$begingroup$
Your inequality is reversed. The correct inequality is
$$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.
As you did, we can show that the inequality above is equivalent to
$$arcsin(b)leq frac{pi}{2},b$$
for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
$$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
by Jensen's Inequality. This shows that
$$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
$endgroup$
add a comment |
$begingroup$
The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
$endgroup$
add a comment |
$begingroup$
The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
$endgroup$
The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
answered Dec 11 '18 at 18:21
Julián AguirreJulián Aguirre
68.6k24094
68.6k24094
add a comment |
add a comment |
$begingroup$
I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
$endgroup$
I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
edited Dec 11 '18 at 18:12
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
answered Dec 11 '18 at 18:03
Sameer BahetiSameer Baheti
5168
5168
add a comment |
add a comment |
$begingroup$
The inequality you seek to prove is not true. Put $b=1/4$.
However, $sin^{-1}ble bpi/2, forall bin[0,1)$
Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$
$f'(x)=pi/2-frac1{sqrt{1-x^2}}$
$f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$
This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
$endgroup$
add a comment |
$begingroup$
The inequality you seek to prove is not true. Put $b=1/4$.
However, $sin^{-1}ble bpi/2, forall bin[0,1)$
Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$
$f'(x)=pi/2-frac1{sqrt{1-x^2}}$
$f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$
This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
$endgroup$
add a comment |
$begingroup$
The inequality you seek to prove is not true. Put $b=1/4$.
However, $sin^{-1}ble bpi/2, forall bin[0,1)$
Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$
$f'(x)=pi/2-frac1{sqrt{1-x^2}}$
$f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$
This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
$endgroup$
The inequality you seek to prove is not true. Put $b=1/4$.
However, $sin^{-1}ble bpi/2, forall bin[0,1)$
Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$
$f'(x)=pi/2-frac1{sqrt{1-x^2}}$
$f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$
This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
edited Dec 11 '18 at 18:29
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
answered Dec 11 '18 at 17:57
Shubham JohriShubham Johri
5,092717
5,092717
add a comment |
add a comment |
$begingroup$
If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$ This implies
$$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$
for $bin (0,1).$ Apply this with $f(b)=arcsin b.$
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
$endgroup$
add a comment |
$begingroup$
If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$ This implies
$$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$
for $bin (0,1).$ Apply this with $f(b)=arcsin b.$
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
$endgroup$
add a comment |
$begingroup$
If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$ This implies
$$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$
for $bin (0,1).$ Apply this with $f(b)=arcsin b.$
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
$endgroup$
If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$ This implies
$$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$
for $bin (0,1).$ Apply this with $f(b)=arcsin b.$
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
answered Dec 11 '18 at 19:29
zhw.zhw.
72.6k43175
72.6k43175
add a comment |
add a comment |
$begingroup$
Your inequality is reversed. The correct inequality is
$$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.
As you did, we can show that the inequality above is equivalent to
$$arcsin(b)leq frac{pi}{2},b$$
for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
$$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
by Jensen's Inequality. This shows that
$$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Your inequality is reversed. The correct inequality is
$$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.
As you did, we can show that the inequality above is equivalent to
$$arcsin(b)leq frac{pi}{2},b$$
for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
$$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
by Jensen's Inequality. This shows that
$$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
$endgroup$
add a comment |
$begingroup$
Your inequality is reversed. The correct inequality is
$$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.
As you did, we can show that the inequality above is equivalent to
$$arcsin(b)leq frac{pi}{2},b$$
for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
$$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
by Jensen's Inequality. This shows that
$$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
$endgroup$
Your inequality is reversed. The correct inequality is
$$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.
As you did, we can show that the inequality above is equivalent to
$$arcsin(b)leq frac{pi}{2},b$$
for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
$$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
by Jensen's Inequality. This shows that
$$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
answered Dec 11 '18 at 23:26
BatominovskiBatominovski
1
1
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$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– Zachary Selk
Dec 11 '18 at 17:51
$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07