Please show $|sin(n+1)x| = |sin(nx+x)|$ [closed]
$begingroup$
Pretty straightforward. I'm looking at a textbook for analysis, in chapter 1, I am not familiar with how
$|sin(n+1)x| = |sin(nx+x)|$
Could someone show me the proof or method here? I'm sure it's quite simple but I don't see it.
trigonometry
edited Dec 19 '13 at 22:17
Namaste
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
$endgroup$
closed as off-topic by José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost, Gibbs Dec 30 '18 at 22:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Pretty straightforward. I'm looking at a textbook for analysis, in chapter 1, I am not familiar with how
$|sin(n+1)x| = |sin(nx+x)|$
Could someone show me the proof or method here? I'm sure it's quite simple but I don't see it.
trigonometry
edited Dec 19 '13 at 22:17
Namaste
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
$endgroup$
closed as off-topic by José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost, Gibbs Dec 30 '18 at 22:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30
add a comment |
$begingroup$
Pretty straightforward. I'm looking at a textbook for analysis, in chapter 1, I am not familiar with how
$|sin(n+1)x| = |sin(nx+x)|$
Could someone show me the proof or method here? I'm sure it's quite simple but I don't see it.
trigonometry
edited Dec 19 '13 at 22:17
Namaste
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
$endgroup$
Pretty straightforward. I'm looking at a textbook for analysis, in chapter 1, I am not familiar with how
$|sin(n+1)x| = |sin(nx+x)|$
Could someone show me the proof or method here? I'm sure it's quite simple but I don't see it.
trigonometry
trigonometry
edited Dec 19 '13 at 22:17
Namaste
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
edited Dec 19 '13 at 22:17
Namaste
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
edited Dec 19 '13 at 22:17
Namaste
1
edited Dec 19 '13 at 22:17
Namaste
1
edited Dec 19 '13 at 22:17
Namaste
1
1
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
asked Dec 19 '13 at 22:12
mathjacksmathjacks
1,68311429
1,68311429
closed as off-topic by José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost, Gibbs Dec 30 '18 at 22:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost, Gibbs Dec 30 '18 at 22:00
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Abcd, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30
add a comment |
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I presume that on the left-hand side, we actually have $|sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$.
It's as simple as that.
answered Dec 19 '13 at 22:13
NamasteNamaste
1
$endgroup$
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
|
show 10 more comments
$begingroup$
Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $sin$, $exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$sin x+sin y =2sintfrac12(x+y)sintfrac12(x-y),$$$$ln xy=ln x+ln y,qquadquad$$$$exp(x+y)=exp x+exp y.qquadqquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$sin nx+x,$$which is the sum of $sin nx$ and $x$.
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I presume that on the left-hand side, we actually have $|sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$.
It's as simple as that.
answered Dec 19 '13 at 22:13
NamasteNamaste
1
$endgroup$
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
|
show 10 more comments
$begingroup$
I presume that on the left-hand side, we actually have $|sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$.
It's as simple as that.
answered Dec 19 '13 at 22:13
NamasteNamaste
1
$endgroup$
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
|
show 10 more comments
$begingroup$
I presume that on the left-hand side, we actually have $|sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$.
It's as simple as that.
answered Dec 19 '13 at 22:13
NamasteNamaste
1
$endgroup$
I presume that on the left-hand side, we actually have $|sin[(n + 1)x]|$, in which case we need only distribute $x$ within the argument of $sin$ to obtain the right-hand side of the equation.
$((n + 1)x) = (nx + x)$.
It's as simple as that.
answered Dec 19 '13 at 22:13
NamasteNamaste
1
edited Jan 12 '14 at 14:14
answered Dec 19 '13 at 22:13
NamasteNamaste
1
answered Dec 19 '13 at 22:13
NamasteNamaste
1
answered Dec 19 '13 at 22:13
NamasteNamaste
1
1
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
|
show 10 more comments
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
7
7
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
$begingroup$
@AtahualpaInca But it answers the question, that's all there is to it. It doesn't make sense to have a large answer on something this simple.
$endgroup$
– Rivasa
Dec 19 '13 at 22:21
3
3
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
$begingroup$
@Atahualpa Before judging, please search the meta.mse site for discussions about the unfortunate growing number of "unanswered" questions, in part due to those who comment but do not answer.
$endgroup$
– Namaste
Dec 19 '13 at 22:25
2
2
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
$begingroup$
@AtahualpaInca I don't dine on questions, sorry about that. Answering questions, at least for me, is not a matter of "competing" or "hunting" or "consuming". It's about helping, at least for me. You're going to make yourself very unhappy, and quickly so, if you view this site as a "game" or competition.
$endgroup$
– Namaste
Dec 19 '13 at 22:27
2
2
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
$begingroup$
@AtahualpaInca I am curious, what do you think should be placed as an answer to a question like this? amWhy's answer seems to be exactly answering the question -- the entire point of an answer.
$endgroup$
– Deven Ware
Dec 19 '13 at 22:46
2
2
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
$begingroup$
@AtahualpaInca There have been numerous discussions about answering questions through comments; the purpose of comments are for clarification and short points, and should be considered ephemeral. In fact, it's discouraged to simply answer in a comment. Regardless, if you object to people giving short and brief answers to questions, then you ought to bring it up on meta, not here.
$endgroup$
– user61527
Dec 19 '13 at 22:50
|
show 10 more comments
$begingroup$
Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $sin$, $exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$sin x+sin y =2sintfrac12(x+y)sintfrac12(x-y),$$$$ln xy=ln x+ln y,qquadquad$$$$exp(x+y)=exp x+exp y.qquadqquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$sin nx+x,$$which is the sum of $sin nx$ and $x$.
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
$endgroup$
add a comment |
$begingroup$
Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $sin$, $exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$sin x+sin y =2sintfrac12(x+y)sintfrac12(x-y),$$$$ln xy=ln x+ln y,qquadquad$$$$exp(x+y)=exp x+exp y.qquadqquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$sin nx+x,$$which is the sum of $sin nx$ and $x$.
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
$endgroup$
add a comment |
$begingroup$
Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $sin$, $exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$sin x+sin y =2sintfrac12(x+y)sintfrac12(x-y),$$$$ln xy=ln x+ln y,qquadquad$$$$exp(x+y)=exp x+exp y.qquadqquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$sin nx+x,$$which is the sum of $sin nx$ and $x$.
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
$endgroup$
Just like the spacing between words in written text, the spacing around binary mathematical operators (such as $+$ or $-$) and unary operators (e.g. $sin$, $exp$, etc.) is critical in determining the meaning.
As a general rule of mathematical typography, mathematical functions written as operators—that is, simply prefixed to the argument—are printed in roman type, followed by a thin space that precedes the argument. For example:$$sin x+sin y =2sintfrac12(x+y)sintfrac12(x-y),$$$$ln xy=ln x+ln y,qquadquad$$$$exp(x+y)=exp x+exp y.qquadqquad$$In the above examples, the typography makes the expressions completely unambiguous. Any additional parentheses would just add clutter, while the existing parentheses are needed.
In the particular case of your question, we have the simple algebraic identity$$(n+1)x=nx+x.$$When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$sin nx+x,$$which is the sum of $sin nx$ and $x$.
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
answered Dec 30 '18 at 14:53
John BentinJohn Bentin
11.4k22554
11.4k22554
add a comment |
add a comment |
$begingroup$
your notation is ambiguous. It looks like your left hand side should be "$sin((n+1)x)$".
$endgroup$
– Stefan Smith
Dec 20 '13 at 3:53
$begingroup$
Pretty straight forward identity.
$endgroup$
– Narasimham
Dec 30 '18 at 20:30