Proving $1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$, for $vartheta in [0, frac{pi}{2}]$
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For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
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add a comment |
$begingroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
$endgroup$
add a comment |
$begingroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
$endgroup$
For my thesis I need the inequality $1 - frac{2 vartheta}{pi} sin vartheta leq 2 cos vartheta$ for $vartheta in [0, frac{pi}{2}]$ which can be proved by exploiting the fact that $cos vartheta$ is concave on $[0, frac{pi}{2}]$.
When I plotted the graph of $1 - frac{2vartheta}{pi} sin vartheta$ and the graph of $cos vartheta$ I noticed that indeed the stronger inequality
$$1 - frac{2 vartheta}{pi} sin vartheta leq cos vartheta$$ seems to hold.
Actually I do not need this stronger version but I would be interested in a proof anyway.
What I have tried:
- Trying to find the zeros of $h(vartheta)=cos vartheta - 1 + frac{2 vartheta}{pi} sin vartheta$.
- Writing down the Taylor-expansion of $h(vartheta)$ and comparing the positive and the negative terms.
Both approaches ended up in a mess. Does anyone have an idea on how to prove this?
trigonometry inequality
trigonometry inequality
edited Dec 14 '18 at 10:51
Blue
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
edited Dec 14 '18 at 10:51
Blue
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
edited Dec 14 '18 at 10:51
Blue
47.7k870151
edited Dec 14 '18 at 10:51
Blue
47.7k870151
edited Dec 14 '18 at 10:51
Blue
47.7k870151
47.7k870151
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
asked Dec 14 '18 at 10:28
Bruno KramsBruno Krams
476
476
add a comment |
add a comment |
1 Answer
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The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
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$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
$begingroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
$endgroup$
The inequality obviously holds for $θ = 0$. For $θ in left(0, dfrac{π}{2}right]$, note thatbegin{align*}
&mathrel{phantom{Longleftrightarrow}}{} 1 - frac{2}{π} θ sin θ leqslant cos θ\
&Longleftrightarrow 2sin^2 frac{θ}{2} = 1 - cos θ leqslant frac{2}{π} θ sin θ = frac{4}{π} θ sinfrac{θ}{2} cosfrac{θ}{2}\
&Longleftrightarrow frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant frac{4}{π}.
end{align*}
Define $f(t) = dfrac{tan t}{t}$ for $t in left(0, dfrac{π}{4}right]$, then $f'(t) = dfrac{2t - sin 2t}{2t^2 cos^2 t} geqslant 0$, which implies that$$
frac{tan dfrac{θ}{2}}{dfrac{θ}{2}} leqslant fleft( frac{π}{4} right) = frac{4}{π}.
$$
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
answered Dec 14 '18 at 10:45
SaadSaad
19.7k92352
19.7k92352
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
1
1
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
$begingroup$
Wow. That was fast and I'm even able to follow your proof :). Thanks a lot.
$endgroup$
– Bruno Krams
Dec 14 '18 at 11:03
add a comment |
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