Questions about Inverse trigonometry functions
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Please note question number 3 has summation from $1$ to $2n$.
This is what I tried for question number $2$.
If I put $n=1$ , the summation for two $arcsin = pi$. This can only happen if both $arcsin=pi/2$ $implies$ $ x_1 = x_2 =1$. Which would mean summation of $x$ from $1$ to $2n$ should be equal to $2n$. The answer given is $0$. Where am I wrong?
I couldn't understand where to start question 3 and 4. Some hints would go a long way. Thanks.
trigonometry
edited Sep 9 at 7:10
Henrik
5,83571930
asked Sep 9 at 6:44
user3508140
19226
 |Â
show 1 more comment
up vote
0
down vote
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Please note question number 3 has summation from $1$ to $2n$.
This is what I tried for question number $2$.
If I put $n=1$ , the summation for two $arcsin = pi$. This can only happen if both $arcsin=pi/2$ $implies$ $ x_1 = x_2 =1$. Which would mean summation of $x$ from $1$ to $2n$ should be equal to $2n$. The answer given is $0$. Where am I wrong?
I couldn't understand where to start question 3 and 4. Some hints would go a long way. Thanks.
trigonometry
edited Sep 9 at 7:10
Henrik
5,83571930
asked Sep 9 at 6:44
user3508140
19226
You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
1
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Please note question number 3 has summation from $1$ to $2n$.
This is what I tried for question number $2$.
If I put $n=1$ , the summation for two $arcsin = pi$. This can only happen if both $arcsin=pi/2$ $implies$ $ x_1 = x_2 =1$. Which would mean summation of $x$ from $1$ to $2n$ should be equal to $2n$. The answer given is $0$. Where am I wrong?
I couldn't understand where to start question 3 and 4. Some hints would go a long way. Thanks.
trigonometry
edited Sep 9 at 7:10
Henrik
5,83571930
asked Sep 9 at 6:44
user3508140
19226
Please note question number 3 has summation from $1$ to $2n$.
This is what I tried for question number $2$.
If I put $n=1$ , the summation for two $arcsin = pi$. This can only happen if both $arcsin=pi/2$ $implies$ $ x_1 = x_2 =1$. Which would mean summation of $x$ from $1$ to $2n$ should be equal to $2n$. The answer given is $0$. Where am I wrong?
I couldn't understand where to start question 3 and 4. Some hints would go a long way. Thanks.
trigonometry
trigonometry
edited Sep 9 at 7:10
Henrik
5,83571930
asked Sep 9 at 6:44
user3508140
19226
edited Sep 9 at 7:10
Henrik
5,83571930
asked Sep 9 at 6:44
user3508140
19226
edited Sep 9 at 7:10
Henrik
5,83571930
edited Sep 9 at 7:10
Henrik
5,83571930
edited Sep 9 at 7:10
Henrik
5,83571930
5,83571930
asked Sep 9 at 6:44
user3508140
19226
asked Sep 9 at 6:44
user3508140
19226
asked Sep 9 at 6:44
user3508140
19226
19226
You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
1
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25
 |Â
show 1 more comment
You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
1
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25
You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
1
1
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25
 |Â
show 1 more comment
1 Answer
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All three questions are based on the same concept of maximum value of $arcsin x$ and $arccos x$ which are $pi/2$ and $pi$.
In three, for example, the maximum value of left side can be $npi$ which is in right side, hence all $x_1,x_2,cdots$ are $1$.
edited Sep 9 at 9:36
Davide Morgante
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
All three questions are based on the same concept of maximum value of $arcsin x$ and $arccos x$ which are $pi/2$ and $pi$.
In three, for example, the maximum value of left side can be $npi$ which is in right side, hence all $x_1,x_2,cdots$ are $1$.
edited Sep 9 at 9:36
Davide Morgante
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
add a comment |Â
up vote
0
down vote
All three questions are based on the same concept of maximum value of $arcsin x$ and $arccos x$ which are $pi/2$ and $pi$.
In three, for example, the maximum value of left side can be $npi$ which is in right side, hence all $x_1,x_2,cdots$ are $1$.
edited Sep 9 at 9:36
Davide Morgante
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
add a comment |Â
up vote
0
down vote
up vote
0
down vote
All three questions are based on the same concept of maximum value of $arcsin x$ and $arccos x$ which are $pi/2$ and $pi$.
In three, for example, the maximum value of left side can be $npi$ which is in right side, hence all $x_1,x_2,cdots$ are $1$.
edited Sep 9 at 9:36
Davide Morgante
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
All three questions are based on the same concept of maximum value of $arcsin x$ and $arccos x$ which are $pi/2$ and $pi$.
In three, for example, the maximum value of left side can be $npi$ which is in right side, hence all $x_1,x_2,cdots$ are $1$.
edited Sep 9 at 9:36
Davide Morgante
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
edited Sep 9 at 9:36
Davide Morgante
2,600724
edited Sep 9 at 9:36
Davide Morgante
2,600724
edited Sep 9 at 9:36
Davide Morgante
2,600724
2,600724
answered Sep 9 at 7:59
ashish deo singh
1556
answered Sep 9 at 7:59
ashish deo singh
1556
answered Sep 9 at 7:59
ashish deo singh
1556
1556
add a comment |Â
add a comment |Â
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You are not wrong. Since $sin^-1xleqfracpi2$, the sum in 2. and 3. equals $npi$ iff all sumands are equal $fracpi2$ iff all $x_i$'s are equal to $1$. So the answer in 2. is $2n$. Also the answer in 3. is $2n(2n-1)$ because you have $(2n)^2$ pairs of $(i,j)$ with $2n$ of them with equal coordinates. For 4. you should note that $cos^-1xleq pi$, hence all summand must be equal to $pi$, i.e. all $x_i$'s are equal to $-1$. Now the sum alternates: $-1+1-1+1-ldots$ and is equal to $-1$ for odd $n$ and $0$ for even $n$.
– SMM
Sep 9 at 6:59
Could you explain question 3 a bit more? @SMM I understand question 4 now.
– user3508140
Sep 9 at 7:15
For question $2$ of your book, the anwser should be $2n$.
– dmtri
Sep 9 at 7:21
1
This sum, as it is written, is over all pairs $(i,j)$ such that $1leq i,jleq 2n$ and $ineq j$. Since you have $2n$ choices for both $i$ and $j$, you have $(2n)^2$ pairs $(i,j)$ such that $1leq i,jleq 2n$. Among these pairs you have $2n$ of them such that $i=j$. So you have $(2n)^2-2n=2n(2n-1)$ pairs satisfying the condition of summation. And each summand is $1$, so the sum is equal to $2n(2n-1)$.
– SMM
Sep 9 at 7:23
For question $3$ of your book the sum should be $n(n-1)=frac2n(n-1)2$ , as it counts double every factor $x_ix_j$
– dmtri
Sep 9 at 7:25