Showing that $I_{n+2}+I_{n}=2I_{n+1}$, where $I_n=int_0^pifrac{1-cos ntheta}{1-costheta}$ [closed]
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I am unable to figure out to how to approach the problem, as all my attempts led to dead ends... Please help me out:
Let $displaystyle I_n =int_0^pi frac{1-cos ntheta}{1-costheta} dtheta$
where $n$ is a positive integer or zero, then show that
$I_{n+2} +I_n =2I_{n+1}$.
Hence prove that
$displaystyle int_0^{pi/2} frac{sin^2 ntheta}{sin^2 theta} dtheta =frac{npi}{2}$.
integration trigonometry trigonometric-integrals
edited Nov 17 at 19:44
Blue
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
closed as off-topic by heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel Nov 18 at 2:53
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
2
down vote
favorite
I am unable to figure out to how to approach the problem, as all my attempts led to dead ends... Please help me out:
Let $displaystyle I_n =int_0^pi frac{1-cos ntheta}{1-costheta} dtheta$
where $n$ is a positive integer or zero, then show that
$I_{n+2} +I_n =2I_{n+1}$.
Hence prove that
$displaystyle int_0^{pi/2} frac{sin^2 ntheta}{sin^2 theta} dtheta =frac{npi}{2}$.
integration trigonometry trigonometric-integrals
edited Nov 17 at 19:44
Blue
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
closed as off-topic by heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel Nov 18 at 2:53
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am unable to figure out to how to approach the problem, as all my attempts led to dead ends... Please help me out:
Let $displaystyle I_n =int_0^pi frac{1-cos ntheta}{1-costheta} dtheta$
where $n$ is a positive integer or zero, then show that
$I_{n+2} +I_n =2I_{n+1}$.
Hence prove that
$displaystyle int_0^{pi/2} frac{sin^2 ntheta}{sin^2 theta} dtheta =frac{npi}{2}$.
integration trigonometry trigonometric-integrals
edited Nov 17 at 19:44
Blue
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
I am unable to figure out to how to approach the problem, as all my attempts led to dead ends... Please help me out:
Let $displaystyle I_n =int_0^pi frac{1-cos ntheta}{1-costheta} dtheta$
where $n$ is a positive integer or zero, then show that
$I_{n+2} +I_n =2I_{n+1}$.
Hence prove that
$displaystyle int_0^{pi/2} frac{sin^2 ntheta}{sin^2 theta} dtheta =frac{npi}{2}$.
integration trigonometry trigonometric-integrals
integration trigonometry trigonometric-integrals
edited Nov 17 at 19:44
Blue
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
edited Nov 17 at 19:44
Blue
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
edited Nov 17 at 19:44
Blue
46.9k870147
edited Nov 17 at 19:44
Blue
46.9k870147
edited Nov 17 at 19:44
Blue
46.9k870147
46.9k870147
asked Nov 17 at 19:13
Abhishek Ghosh
405
asked Nov 17 at 19:13
Abhishek Ghosh
405
asked Nov 17 at 19:13
Abhishek Ghosh
405
405
closed as off-topic by heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel Nov 18 at 2:53
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel Nov 18 at 2:53
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, José Carlos Santos, Nosrati, user10354138, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59
add a comment |
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
hint
Observe that
$$cos((n+2)theta)+cos(ntheta)=$$
$$2cos((n+1)theta)Bigl(cos(theta)-1+1Bigr)$$
and
$$int_0^picos((n+1)theta)dtheta=0$$
we have then
$$I_{n+2}-I_{n+1}=I_{n+1}-I_n=I_1-I_0=pi$$
$(I_n)$ is an arithmetic sequence.
thus
$$I_n=I_0+npi=npi$$
on the other hand, by the substitution
$theta=2u$,
we find that
$$I_n=2int_0^frac{pi}{2}frac{sin^2(nu)}{sin^2(u)}du$$
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
hint
Observe that
$$cos((n+2)theta)+cos(ntheta)=$$
$$2cos((n+1)theta)Bigl(cos(theta)-1+1Bigr)$$
and
$$int_0^picos((n+1)theta)dtheta=0$$
we have then
$$I_{n+2}-I_{n+1}=I_{n+1}-I_n=I_1-I_0=pi$$
$(I_n)$ is an arithmetic sequence.
thus
$$I_n=I_0+npi=npi$$
on the other hand, by the substitution
$theta=2u$,
we find that
$$I_n=2int_0^frac{pi}{2}frac{sin^2(nu)}{sin^2(u)}du$$
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
add a comment |
up vote
4
down vote
accepted
hint
Observe that
$$cos((n+2)theta)+cos(ntheta)=$$
$$2cos((n+1)theta)Bigl(cos(theta)-1+1Bigr)$$
and
$$int_0^picos((n+1)theta)dtheta=0$$
we have then
$$I_{n+2}-I_{n+1}=I_{n+1}-I_n=I_1-I_0=pi$$
$(I_n)$ is an arithmetic sequence.
thus
$$I_n=I_0+npi=npi$$
on the other hand, by the substitution
$theta=2u$,
we find that
$$I_n=2int_0^frac{pi}{2}frac{sin^2(nu)}{sin^2(u)}du$$
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
hint
Observe that
$$cos((n+2)theta)+cos(ntheta)=$$
$$2cos((n+1)theta)Bigl(cos(theta)-1+1Bigr)$$
and
$$int_0^picos((n+1)theta)dtheta=0$$
we have then
$$I_{n+2}-I_{n+1}=I_{n+1}-I_n=I_1-I_0=pi$$
$(I_n)$ is an arithmetic sequence.
thus
$$I_n=I_0+npi=npi$$
on the other hand, by the substitution
$theta=2u$,
we find that
$$I_n=2int_0^frac{pi}{2}frac{sin^2(nu)}{sin^2(u)}du$$
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
hint
Observe that
$$cos((n+2)theta)+cos(ntheta)=$$
$$2cos((n+1)theta)Bigl(cos(theta)-1+1Bigr)$$
and
$$int_0^picos((n+1)theta)dtheta=0$$
we have then
$$I_{n+2}-I_{n+1}=I_{n+1}-I_n=I_1-I_0=pi$$
$(I_n)$ is an arithmetic sequence.
thus
$$I_n=I_0+npi=npi$$
on the other hand, by the substitution
$theta=2u$,
we find that
$$I_n=2int_0^frac{pi}{2}frac{sin^2(nu)}{sin^2(u)}du$$
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
edited Nov 17 at 20:30
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
answered Nov 17 at 19:22
hamam_Abdallah
36.9k21533
36.9k21533
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
add a comment |
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
Even I know this formula... The problem is how to apply it here
– Abhishek Ghosh
Nov 17 at 19:32
2
2
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
@AbhishekGhosh: To avoid people telling you things you already know, you should edit your question to include some of those things you've tried. You mentioned hitting dead ends; show us those ends, and we might be able to get you past them. The more we know about what you know, the better we can tailor responses to your experience level. (Also, providing your thoughts on a problem helps convince people that you aren't merely trying to get them to do your homework for them.)
– Blue
Nov 17 at 19:50
add a comment |
See math.stackexchange.com/questions/2008044/…
– lab bhattacharjee
Nov 18 at 1:57
and math.stackexchange.com/questions/263705/…
– lab bhattacharjee
Nov 18 at 1:59