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Monotonicity of trigonometric function
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up vote
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Consider the function
$$f(x)=fracsin(x)sin((2k+1)x)$$
with $k$ a positive integer. It seems that $f$ is strictly increasing in $[0,fracpi2(2k+1)]$. Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?
functions trigonometry monotone-functions
edited Aug 8 at 11:56
Jam
4,35811330
asked Aug 8 at 11:23
RTJ
3,5762621
add a comment |Â
up vote
8
down vote
favorite
Consider the function
$$f(x)=fracsin(x)sin((2k+1)x)$$
with $k$ a positive integer. It seems that $f$ is strictly increasing in $[0,fracpi2(2k+1)]$. Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?
functions trigonometry monotone-functions
edited Aug 8 at 11:56
Jam
4,35811330
asked Aug 8 at 11:23
RTJ
3,5762621
1
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Consider the function
$$f(x)=fracsin(x)sin((2k+1)x)$$
with $k$ a positive integer. It seems that $f$ is strictly increasing in $[0,fracpi2(2k+1)]$. Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?
functions trigonometry monotone-functions
edited Aug 8 at 11:56
Jam
4,35811330
asked Aug 8 at 11:23
RTJ
3,5762621
Consider the function
$$f(x)=fracsin(x)sin((2k+1)x)$$
with $k$ a positive integer. It seems that $f$ is strictly increasing in $[0,fracpi2(2k+1)]$. Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?
functions trigonometry monotone-functions
edited Aug 8 at 11:56
Jam
4,35811330
asked Aug 8 at 11:23
RTJ
3,5762621
edited Aug 8 at 11:56
Jam
4,35811330
edited Aug 8 at 11:56
Jam
4,35811330
edited Aug 8 at 11:56
Jam
4,35811330
4,35811330
asked Aug 8 at 11:23
RTJ
3,5762621
asked Aug 8 at 11:23
RTJ
3,5762621
asked Aug 8 at 11:23
RTJ
3,5762621
3,5762621
1
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37
add a comment |Â
1
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37
1
1
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$$beginalignedfracsin(x)sin((k+1)x)&=fracsin(x)sin(kx)cos(x)+cos(kx)sin(x)\
&=frac1fracsin(kx)sin(x)cos(x)+cos(kx)
endaligned$$
Hence: $$beginalignedfracsin(x)sin(kx)text increasing&ifffracsin(kx)sin(x)text decreasing\
&Rightarrow fracsin(kx)sin(x)cos(x)text decreasing\
&ifffracsin(kx)sin(x)cos(x)+cos(kx)text decreasing\
&ifffrac1fracsin(kx)sin(x)cos(x)+cos(kx)text increasing\
\
&ifffracsin(x)sin((k+1)x)text increasing\
endaligned$$
Let $P(k)$ denote the proposition that $fracsin(x)sin(kx)$ is increasing. Hence, we've shown that $P(k)rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $fracsin(x)sin((2k+1)x)$ is increasing over $[0,fracpi2(2k+1)]$ for all $kinmathbbN$.
answered Aug 8 at 12:11
Jam
4,35811330
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
 |Â
show 4 more comments
up vote
4
down vote
You have $sin2x=2sin x cos x$ and $sin 3x=sin x cos 2x + cos x sin 2x =sin x (cos 2 x + 2 cos^2 x)$, and similar formulas for higher $sin (kx)$. Then
$$fracsin(3x)sin x = cos 2 x + 2cos^2x
$$
where the RHS is decreasing in $x$ and positive as long as $cos 2x>0$. That gives you an interval where $fracsin xsin 3x$ is increasing.
Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).
answered Aug 8 at 11:44
Kusma
2,567214
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$beginalignedfracsin(x)sin((k+1)x)&=fracsin(x)sin(kx)cos(x)+cos(kx)sin(x)\
&=frac1fracsin(kx)sin(x)cos(x)+cos(kx)
endaligned$$
Hence: $$beginalignedfracsin(x)sin(kx)text increasing&ifffracsin(kx)sin(x)text decreasing\
&Rightarrow fracsin(kx)sin(x)cos(x)text decreasing\
&ifffracsin(kx)sin(x)cos(x)+cos(kx)text decreasing\
&ifffrac1fracsin(kx)sin(x)cos(x)+cos(kx)text increasing\
\
&ifffracsin(x)sin((k+1)x)text increasing\
endaligned$$
Let $P(k)$ denote the proposition that $fracsin(x)sin(kx)$ is increasing. Hence, we've shown that $P(k)rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $fracsin(x)sin((2k+1)x)$ is increasing over $[0,fracpi2(2k+1)]$ for all $kinmathbbN$.
answered Aug 8 at 12:11
Jam
4,35811330
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
 |Â
show 4 more comments
up vote
4
down vote
accepted
$$beginalignedfracsin(x)sin((k+1)x)&=fracsin(x)sin(kx)cos(x)+cos(kx)sin(x)\
&=frac1fracsin(kx)sin(x)cos(x)+cos(kx)
endaligned$$
Hence: $$beginalignedfracsin(x)sin(kx)text increasing&ifffracsin(kx)sin(x)text decreasing\
&Rightarrow fracsin(kx)sin(x)cos(x)text decreasing\
&ifffracsin(kx)sin(x)cos(x)+cos(kx)text decreasing\
&ifffrac1fracsin(kx)sin(x)cos(x)+cos(kx)text increasing\
\
&ifffracsin(x)sin((k+1)x)text increasing\
endaligned$$
Let $P(k)$ denote the proposition that $fracsin(x)sin(kx)$ is increasing. Hence, we've shown that $P(k)rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $fracsin(x)sin((2k+1)x)$ is increasing over $[0,fracpi2(2k+1)]$ for all $kinmathbbN$.
answered Aug 8 at 12:11
Jam
4,35811330
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
 |Â
show 4 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$beginalignedfracsin(x)sin((k+1)x)&=fracsin(x)sin(kx)cos(x)+cos(kx)sin(x)\
&=frac1fracsin(kx)sin(x)cos(x)+cos(kx)
endaligned$$
Hence: $$beginalignedfracsin(x)sin(kx)text increasing&ifffracsin(kx)sin(x)text decreasing\
&Rightarrow fracsin(kx)sin(x)cos(x)text decreasing\
&ifffracsin(kx)sin(x)cos(x)+cos(kx)text decreasing\
&ifffrac1fracsin(kx)sin(x)cos(x)+cos(kx)text increasing\
\
&ifffracsin(x)sin((k+1)x)text increasing\
endaligned$$
Let $P(k)$ denote the proposition that $fracsin(x)sin(kx)$ is increasing. Hence, we've shown that $P(k)rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $fracsin(x)sin((2k+1)x)$ is increasing over $[0,fracpi2(2k+1)]$ for all $kinmathbbN$.
answered Aug 8 at 12:11
Jam
4,35811330
$$beginalignedfracsin(x)sin((k+1)x)&=fracsin(x)sin(kx)cos(x)+cos(kx)sin(x)\
&=frac1fracsin(kx)sin(x)cos(x)+cos(kx)
endaligned$$
Hence: $$beginalignedfracsin(x)sin(kx)text increasing&ifffracsin(kx)sin(x)text decreasing\
&Rightarrow fracsin(kx)sin(x)cos(x)text decreasing\
&ifffracsin(kx)sin(x)cos(x)+cos(kx)text decreasing\
&ifffrac1fracsin(kx)sin(x)cos(x)+cos(kx)text increasing\
\
&ifffracsin(x)sin((k+1)x)text increasing\
endaligned$$
Let $P(k)$ denote the proposition that $fracsin(x)sin(kx)$ is increasing. Hence, we've shown that $P(k)rightarrow P(k+1)$. So, we may use $P(1)$ (as shown in @Kusma's answer) to inductively prove that $fracsin(x)sin((2k+1)x)$ is increasing over $[0,fracpi2(2k+1)]$ for all $kinmathbbN$.
answered Aug 8 at 12:11
Jam
4,35811330
edited Aug 8 at 14:35
answered Aug 8 at 12:11
Jam
4,35811330
answered Aug 8 at 12:11
Jam
4,35811330
answered Aug 8 at 12:11
Jam
4,35811330
4,35811330
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
 |Â
show 4 more comments
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
Thank you! This provides a recursive relation to show that all functions $fracsin(kx)sin(x)$ are decreasing. So it solves the full problem. I will accept this answer! Maybe you could possibly add the details for future reference.
– RTJ
Aug 8 at 12:26
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
@RTJ Which details would you like?
– Jam
Aug 8 at 12:27
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Well, an induction argument can be used i.e. starting from $sin(2x)/sinx$ we can recursively show using your argument that all these functions are decreasing.
– RTJ
Aug 8 at 12:28
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
Sure thing, I'll do that now.
– Jam
Aug 8 at 12:32
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
@RTJ I'm not entirely convinced there's enough to form a full inductive argument since we've shown that $P(2k+1)rightarrow P(2k)$ but I don't think we've proven that $P(2k)rightarrow P(2k+1)$.
– Jam
Aug 8 at 12:46
 |Â
show 4 more comments
up vote
4
down vote
You have $sin2x=2sin x cos x$ and $sin 3x=sin x cos 2x + cos x sin 2x =sin x (cos 2 x + 2 cos^2 x)$, and similar formulas for higher $sin (kx)$. Then
$$fracsin(3x)sin x = cos 2 x + 2cos^2x
$$
where the RHS is decreasing in $x$ and positive as long as $cos 2x>0$. That gives you an interval where $fracsin xsin 3x$ is increasing.
Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).
answered Aug 8 at 11:44
Kusma
2,567214
add a comment |Â
up vote
4
down vote
You have $sin2x=2sin x cos x$ and $sin 3x=sin x cos 2x + cos x sin 2x =sin x (cos 2 x + 2 cos^2 x)$, and similar formulas for higher $sin (kx)$. Then
$$fracsin(3x)sin x = cos 2 x + 2cos^2x
$$
where the RHS is decreasing in $x$ and positive as long as $cos 2x>0$. That gives you an interval where $fracsin xsin 3x$ is increasing.
Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).
answered Aug 8 at 11:44
Kusma
2,567214
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You have $sin2x=2sin x cos x$ and $sin 3x=sin x cos 2x + cos x sin 2x =sin x (cos 2 x + 2 cos^2 x)$, and similar formulas for higher $sin (kx)$. Then
$$fracsin(3x)sin x = cos 2 x + 2cos^2x
$$
where the RHS is decreasing in $x$ and positive as long as $cos 2x>0$. That gives you an interval where $fracsin xsin 3x$ is increasing.
Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).
answered Aug 8 at 11:44
Kusma
2,567214
You have $sin2x=2sin x cos x$ and $sin 3x=sin x cos 2x + cos x sin 2x =sin x (cos 2 x + 2 cos^2 x)$, and similar formulas for higher $sin (kx)$. Then
$$fracsin(3x)sin x = cos 2 x + 2cos^2x
$$
where the RHS is decreasing in $x$ and positive as long as $cos 2x>0$. That gives you an interval where $fracsin xsin 3x$ is increasing.
Of course turning this into a proof for general $k$ is a lot more complicated than differentiating, but it should be possible. (You don't get non-integer $k$, though).
answered Aug 8 at 11:44
Kusma
2,567214
answered Aug 8 at 11:44
Kusma
2,567214
answered Aug 8 at 11:44
Kusma
2,567214
answered Aug 8 at 11:44
Kusma
2,567214
2,567214
add a comment |Â
add a comment |Â
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1
I think you mean that $f$ is strictly increasing in $[0, fracpi2k+1]$. The Desmos graph shows that this is not true for $k=1$.
– Toby Mak
Aug 8 at 11:28
@TobyMak Yes, sorry I will correct the question. I am actually looking for the minimum.
– RTJ
Aug 8 at 11:32
@TobyMak Not up to $pi/(2k+1)$. Up to half the distance $pi/(2(2k+1))$.
– RTJ
Aug 8 at 11:37