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Some questions about trigonometric polynomials, Vieta's relations and correct integer parameters for roots
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I was trying to compute:
$$P=sin(2pi/7) sin(4pi/7) sin(8pi/7)$$
So I managed to analyse the equation $sin(7x)=0$, thus $x = kpi/7$, when $k$ is an integer.
The problem arrives when the equation
$sin(7x) = 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x)=0$ is derived, because I can't exactly know what are its roots.
Assuming $sin(x) ≠0$, the equation becomes
$$7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)= 7 - 56t^2 + 112t^4 - 64t^6 = 0$$
Where $t=sin(x)$.
But, to solve my problem, how can I set the values for the integer parameter $k$, since I've already looked in Wolfram, and the roots of that equation in $t$ are, apparently, $± ,sin(2pi/7),,± ,sin(4pi/7),,± ,sin(8pi/7)$, so some good values for $k$ are $±,1,,±,2$ and $,±,3$, since we have three pairs of roots with opposite signs, and, by setting that, I successfully get the desired result, $P=-,fracsqrt78$.
But, I don't know how to set those "good" values for $k$, can you help me?
Another question, that has to do with this one, is it correct to affirm that the Vieta's relation does not hold for trigonometric polynomials like the one for $fracsin(7x)sin(x)=7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)$, because the sign of the roots would not satisfy it?
Thanks in advance!
trigonometry polynomials roots products
edited Jul 16 at 13:48
GuySa
402313
asked Jul 16 at 5:11
BrazilianAeronautics
5615
add a comment |Â
up vote
2
down vote
favorite
I was trying to compute:
$$P=sin(2pi/7) sin(4pi/7) sin(8pi/7)$$
So I managed to analyse the equation $sin(7x)=0$, thus $x = kpi/7$, when $k$ is an integer.
The problem arrives when the equation
$sin(7x) = 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x)=0$ is derived, because I can't exactly know what are its roots.
Assuming $sin(x) ≠0$, the equation becomes
$$7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)= 7 - 56t^2 + 112t^4 - 64t^6 = 0$$
Where $t=sin(x)$.
But, to solve my problem, how can I set the values for the integer parameter $k$, since I've already looked in Wolfram, and the roots of that equation in $t$ are, apparently, $± ,sin(2pi/7),,± ,sin(4pi/7),,± ,sin(8pi/7)$, so some good values for $k$ are $±,1,,±,2$ and $,±,3$, since we have three pairs of roots with opposite signs, and, by setting that, I successfully get the desired result, $P=-,fracsqrt78$.
But, I don't know how to set those "good" values for $k$, can you help me?
Another question, that has to do with this one, is it correct to affirm that the Vieta's relation does not hold for trigonometric polynomials like the one for $fracsin(7x)sin(x)=7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)$, because the sign of the roots would not satisfy it?
Thanks in advance!
trigonometry polynomials roots products
edited Jul 16 at 13:48
GuySa
402313
asked Jul 16 at 5:11
BrazilianAeronautics
5615
math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was trying to compute:
$$P=sin(2pi/7) sin(4pi/7) sin(8pi/7)$$
So I managed to analyse the equation $sin(7x)=0$, thus $x = kpi/7$, when $k$ is an integer.
The problem arrives when the equation
$sin(7x) = 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x)=0$ is derived, because I can't exactly know what are its roots.
Assuming $sin(x) ≠0$, the equation becomes
$$7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)= 7 - 56t^2 + 112t^4 - 64t^6 = 0$$
Where $t=sin(x)$.
But, to solve my problem, how can I set the values for the integer parameter $k$, since I've already looked in Wolfram, and the roots of that equation in $t$ are, apparently, $± ,sin(2pi/7),,± ,sin(4pi/7),,± ,sin(8pi/7)$, so some good values for $k$ are $±,1,,±,2$ and $,±,3$, since we have three pairs of roots with opposite signs, and, by setting that, I successfully get the desired result, $P=-,fracsqrt78$.
But, I don't know how to set those "good" values for $k$, can you help me?
Another question, that has to do with this one, is it correct to affirm that the Vieta's relation does not hold for trigonometric polynomials like the one for $fracsin(7x)sin(x)=7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)$, because the sign of the roots would not satisfy it?
Thanks in advance!
trigonometry polynomials roots products
edited Jul 16 at 13:48
GuySa
402313
asked Jul 16 at 5:11
BrazilianAeronautics
5615
I was trying to compute:
$$P=sin(2pi/7) sin(4pi/7) sin(8pi/7)$$
So I managed to analyse the equation $sin(7x)=0$, thus $x = kpi/7$, when $k$ is an integer.
The problem arrives when the equation
$sin(7x) = 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x)=0$ is derived, because I can't exactly know what are its roots.
Assuming $sin(x) ≠0$, the equation becomes
$$7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)= 7 - 56t^2 + 112t^4 - 64t^6 = 0$$
Where $t=sin(x)$.
But, to solve my problem, how can I set the values for the integer parameter $k$, since I've already looked in Wolfram, and the roots of that equation in $t$ are, apparently, $± ,sin(2pi/7),,± ,sin(4pi/7),,± ,sin(8pi/7)$, so some good values for $k$ are $±,1,,±,2$ and $,±,3$, since we have three pairs of roots with opposite signs, and, by setting that, I successfully get the desired result, $P=-,fracsqrt78$.
But, I don't know how to set those "good" values for $k$, can you help me?
Another question, that has to do with this one, is it correct to affirm that the Vieta's relation does not hold for trigonometric polynomials like the one for $fracsin(7x)sin(x)=7 - 56sin^2(x) + 112sin^4(x) - 64sin^6(x)$, because the sign of the roots would not satisfy it?
Thanks in advance!
trigonometry polynomials roots products
edited Jul 16 at 13:48
GuySa
402313
asked Jul 16 at 5:11
BrazilianAeronautics
5615
edited Jul 16 at 13:48
GuySa
402313
edited Jul 16 at 13:48
GuySa
402313
edited Jul 16 at 13:48
GuySa
402313
402313
asked Jul 16 at 5:11
BrazilianAeronautics
5615
asked Jul 16 at 5:11
BrazilianAeronautics
5615
asked Jul 16 at 5:11
BrazilianAeronautics
5615
5615
math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40
add a comment |Â
math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40
math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40
math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40
add a comment |Â
1 Answer
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You have
$$fracsin 7xsin x=P(sin x)$$
where
$$P(t)=7-56t^2+112t^4-64t^6.$$
The distinct roots of $P(t)$ are $pmsin(2pi/7)$, $pmsin(4pi/7)$
and $pmsin(6pi/7)$. But $P(t)=Q(t^2)$ where
$$Q(u)=7-56u+112u^2-64u^3.$$
The distinct roots of $Q(u)$ are $sin^2(2pi/7)$, $sin^2(4pi/7)$
and $sin^2(6pi/7)$. By Vieta,
$$sin^2(2pi/7)sin^2(4pi/7)sin^2(6pi/7)=frac764.$$
Taking square roots:
$$sin(2pi/7)sin(4pi/7)sin(6pi/7)=fracsqrt78$$
as these sines are all positive. Finally
$$sin(2pi/7)sin(4pi/7)sin(8pi/7)=-fracsqrt78$$
as $sin(8pi/7)=-sin(6pi/7)$.
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
You have
$$fracsin 7xsin x=P(sin x)$$
where
$$P(t)=7-56t^2+112t^4-64t^6.$$
The distinct roots of $P(t)$ are $pmsin(2pi/7)$, $pmsin(4pi/7)$
and $pmsin(6pi/7)$. But $P(t)=Q(t^2)$ where
$$Q(u)=7-56u+112u^2-64u^3.$$
The distinct roots of $Q(u)$ are $sin^2(2pi/7)$, $sin^2(4pi/7)$
and $sin^2(6pi/7)$. By Vieta,
$$sin^2(2pi/7)sin^2(4pi/7)sin^2(6pi/7)=frac764.$$
Taking square roots:
$$sin(2pi/7)sin(4pi/7)sin(6pi/7)=fracsqrt78$$
as these sines are all positive. Finally
$$sin(2pi/7)sin(4pi/7)sin(8pi/7)=-fracsqrt78$$
as $sin(8pi/7)=-sin(6pi/7)$.
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
5
down vote
You have
$$fracsin 7xsin x=P(sin x)$$
where
$$P(t)=7-56t^2+112t^4-64t^6.$$
The distinct roots of $P(t)$ are $pmsin(2pi/7)$, $pmsin(4pi/7)$
and $pmsin(6pi/7)$. But $P(t)=Q(t^2)$ where
$$Q(u)=7-56u+112u^2-64u^3.$$
The distinct roots of $Q(u)$ are $sin^2(2pi/7)$, $sin^2(4pi/7)$
and $sin^2(6pi/7)$. By Vieta,
$$sin^2(2pi/7)sin^2(4pi/7)sin^2(6pi/7)=frac764.$$
Taking square roots:
$$sin(2pi/7)sin(4pi/7)sin(6pi/7)=fracsqrt78$$
as these sines are all positive. Finally
$$sin(2pi/7)sin(4pi/7)sin(8pi/7)=-fracsqrt78$$
as $sin(8pi/7)=-sin(6pi/7)$.
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You have
$$fracsin 7xsin x=P(sin x)$$
where
$$P(t)=7-56t^2+112t^4-64t^6.$$
The distinct roots of $P(t)$ are $pmsin(2pi/7)$, $pmsin(4pi/7)$
and $pmsin(6pi/7)$. But $P(t)=Q(t^2)$ where
$$Q(u)=7-56u+112u^2-64u^3.$$
The distinct roots of $Q(u)$ are $sin^2(2pi/7)$, $sin^2(4pi/7)$
and $sin^2(6pi/7)$. By Vieta,
$$sin^2(2pi/7)sin^2(4pi/7)sin^2(6pi/7)=frac764.$$
Taking square roots:
$$sin(2pi/7)sin(4pi/7)sin(6pi/7)=fracsqrt78$$
as these sines are all positive. Finally
$$sin(2pi/7)sin(4pi/7)sin(8pi/7)=-fracsqrt78$$
as $sin(8pi/7)=-sin(6pi/7)$.
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
You have
$$fracsin 7xsin x=P(sin x)$$
where
$$P(t)=7-56t^2+112t^4-64t^6.$$
The distinct roots of $P(t)$ are $pmsin(2pi/7)$, $pmsin(4pi/7)$
and $pmsin(6pi/7)$. But $P(t)=Q(t^2)$ where
$$Q(u)=7-56u+112u^2-64u^3.$$
The distinct roots of $Q(u)$ are $sin^2(2pi/7)$, $sin^2(4pi/7)$
and $sin^2(6pi/7)$. By Vieta,
$$sin^2(2pi/7)sin^2(4pi/7)sin^2(6pi/7)=frac764.$$
Taking square roots:
$$sin(2pi/7)sin(4pi/7)sin(6pi/7)=fracsqrt78$$
as these sines are all positive. Finally
$$sin(2pi/7)sin(4pi/7)sin(8pi/7)=-fracsqrt78$$
as $sin(8pi/7)=-sin(6pi/7)$.
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
edited Jul 16 at 5:59
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
answered Jul 16 at 5:39
Lord Shark the Unknown
85.7k951112
85.7k951112
add a comment |Â
add a comment |Â
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math.stackexchange.com/questions/311781/…
– lab bhattacharjee
Jul 16 at 8:40