How to convert $1<frac{x}{sin x}frac{sin x}{x}> cos x$?
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How to convert $1<frac{x}{sin x}<frac{1}{cos x}$ to $1>frac{sin x}{x}> cos x$?
This is done in the process of deriving the sine function. This is maybe a simple but important bug in my understanding that needs to be fixed. Thanks in advance
calculus real-analysis limits
edited Nov 26 at 16:23
Shaun
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
add a comment |
up vote
0
down vote
favorite
How to convert $1<frac{x}{sin x}<frac{1}{cos x}$ to $1>frac{sin x}{x}> cos x$?
This is done in the process of deriving the sine function. This is maybe a simple but important bug in my understanding that needs to be fixed. Thanks in advance
calculus real-analysis limits
edited Nov 26 at 16:23
Shaun
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
3
If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
2
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to convert $1<frac{x}{sin x}<frac{1}{cos x}$ to $1>frac{sin x}{x}> cos x$?
This is done in the process of deriving the sine function. This is maybe a simple but important bug in my understanding that needs to be fixed. Thanks in advance
calculus real-analysis limits
edited Nov 26 at 16:23
Shaun
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
How to convert $1<frac{x}{sin x}<frac{1}{cos x}$ to $1>frac{sin x}{x}> cos x$?
This is done in the process of deriving the sine function. This is maybe a simple but important bug in my understanding that needs to be fixed. Thanks in advance
calculus real-analysis limits
calculus real-analysis limits
edited Nov 26 at 16:23
Shaun
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
edited Nov 26 at 16:23
Shaun
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
edited Nov 26 at 16:23
Shaun
8,190113577
edited Nov 26 at 16:23
Shaun
8,190113577
edited Nov 26 at 16:23
Shaun
8,190113577
8,190113577
asked Jun 18 '15 at 19:31
Zedrijn
1448
asked Jun 18 '15 at 19:31
Zedrijn
1448
asked Jun 18 '15 at 19:31
Zedrijn
1448
1448
3
If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
2
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00
add a comment |
3
If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
2
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00
3
3
If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
2
2
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00
add a comment |
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If $0 < a < b$, then $0 < frac{1}{b} < frac{1}{a}$, for $$frac{1}{a} - frac{1}{b} = frac{b-a}{ab},$$ and by assumption $b-a > 0$ and $ab > 0$.
– Daniel Fischer♦
Jun 18 '15 at 19:33
This means $1<frac{x}{sinx}$ and $frac{x}{sinx}<frac{1}{cosx}$. Now multiplication and division (assuming x, $sinx$ and $cosx$ are all positive) easily makes the first inequality equivalent to $frac{sinx}{x}<1$, which is your first requirement. Similarly for the second inequality.
– Paul
Jun 18 '15 at 20:49
2
@Paul But you don't need all of $x, sin x, cos x$ to be positive (they almost certainly aren't in the context of differentiation where $x$ likely approaches $0$ from either side). It's enough for $cos x$ and $sin x/x$ to be positive.
– Erick Wong
Jun 21 '15 at 18:00