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Trigonometry : Find the value of $csc^2 pi/7 + csc^2 2pi/7 + csc^2 3pi/7$ [duplicate]
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Find the value of $textrmcosec^2left(fracpi7right) +textrmcosec^2left(frac2pi7right)+textrmcosec^2left(frac4pi7right)$
7 answers
Find the value of $csc^2 pi/7 + csc^2 2pi/7 + csc^2 3pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
trigonometry
edited Jul 30 at 15:17
amWhy
189k25219431
asked Jul 30 at 15:04
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marked as duplicate by Batominovski, N. F. Taussig, lab bhattacharjee trigonometry
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Jul 30 at 17:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Find the value of $textrmcosec^2left(fracpi7right) +textrmcosec^2left(frac2pi7right)+textrmcosec^2left(frac4pi7right)$
7 answers
Find the value of $csc^2 pi/7 + csc^2 2pi/7 + csc^2 3pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
trigonometry
edited Jul 30 at 15:17
amWhy
189k25219431
asked Jul 30 at 15:04
user580093
204
marked as duplicate by Batominovski, N. F. Taussig, lab bhattacharjee trigonometry
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Jul 30 at 17:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57
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up vote
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This question already has an answer here:
Find the value of $textrmcosec^2left(fracpi7right) +textrmcosec^2left(frac2pi7right)+textrmcosec^2left(frac4pi7right)$
7 answers
Find the value of $csc^2 pi/7 + csc^2 2pi/7 + csc^2 3pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
trigonometry
edited Jul 30 at 15:17
amWhy
189k25219431
asked Jul 30 at 15:04
user580093
204
This question already has an answer here:
Find the value of $textrmcosec^2left(fracpi7right) +textrmcosec^2left(frac2pi7right)+textrmcosec^2left(frac4pi7right)$
7 answers
Find the value of $csc^2 pi/7 + csc^2 2pi/7 + csc^2 3pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
This question already has an answer here:
Find the value of $textrmcosec^2left(fracpi7right) +textrmcosec^2left(frac2pi7right)+textrmcosec^2left(frac4pi7right)$
7 answers
trigonometry
edited Jul 30 at 15:17
amWhy
189k25219431
asked Jul 30 at 15:04
user580093
204
edited Jul 30 at 15:17
amWhy
189k25219431
edited Jul 30 at 15:17
amWhy
189k25219431
edited Jul 30 at 15:17
amWhy
189k25219431
189k25219431
asked Jul 30 at 15:04
user580093
204
asked Jul 30 at 15:04
user580093
204
asked Jul 30 at 15:04
user580093
204
204
marked as duplicate by Batominovski, N. F. Taussig, lab bhattacharjee trigonometry
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marked as duplicate by Batominovski, N. F. Taussig, lab bhattacharjee trigonometry
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Jul 30 at 17:15
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57
add a comment |Â
1
Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57
1
1
Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57
add a comment |Â
2 Answers
2
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votes
up vote
0
down vote
The same question is answered here with $csc^2(4pi/7)$ instead of $csc^2(3pi/7)$.
answered Jul 30 at 15:22
Lod
286
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
add a comment |Â
up vote
0
down vote
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $tan$ and $cot$ need to reinforce.
First, some preparation.
We all know that
$$
tan (x pm y) = frac tan (x) pm tan (y) 1 mp tan (x)tan (y).
$$
Then
$$
cot (x - y) = frac cot (x) cot (y) + 1 cot (y) - cot (x),
$$
or
$$
cot (x) cot (y) = cot(x-y) (cot(y) -cot(x)) - 1
$$
Especially,
$$
cot (2x) = frac cot^2 (x) -1 2 cot (x),
$$
which yields
$$
cot^2(x) - 1 = 2cot(x)cot(2x).
$$
Also,
$$
csc^2(x) = frac sin^2(x) + cos^2(x) sin ^2(x) = 1 + cot^2(x).
$$
Now let's start. Let $x = pi /7$.
$$
P := csc^2(x) + csc^2(2x) + csc^2(4x) = 3 + cot^2(x) + cot^2(2x) + cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(cot(x) cot(2x) + cot(2x) cot (4x) + cot (4x) cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $cot(pi pm y) = mpcot (y)$, we have
$$
Q = cot(x)cot(2x) - cot(2x) cot (3x) - cot(3x) cot (x).
$$
Now,
beginalign*
Q &= 1+ cot(x)(cot(x) - cot(2x)) - cot(x) (cot(2x) -cot(3x)) - cot(2x)(cot(x) - cot (3x))\
&= 1+ cot(x) (cot(x) -3cot(2x) + cot(3x)) + cot(2x) cot(3x)\
&= cot(x) (cot(x) -3cot(2x) + cot(3x)) + (cot(2x)-cot(3x))cot(x) \
&= cot(x)(cot(x) -2cot(2x))\
&= cot^2(x) - 2cot(2x) cot(x)\
&=1.
endalign*
Therefore $P = 6+2Q=8$.
answered Jul 30 at 17:19
xbh
1,0257
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The same question is answered here with $csc^2(4pi/7)$ instead of $csc^2(3pi/7)$.
answered Jul 30 at 15:22
Lod
286
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
add a comment |Â
up vote
0
down vote
The same question is answered here with $csc^2(4pi/7)$ instead of $csc^2(3pi/7)$.
answered Jul 30 at 15:22
Lod
286
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The same question is answered here with $csc^2(4pi/7)$ instead of $csc^2(3pi/7)$.
answered Jul 30 at 15:22
Lod
286
The same question is answered here with $csc^2(4pi/7)$ instead of $csc^2(3pi/7)$.
answered Jul 30 at 15:22
Lod
286
edited Jul 30 at 15:34
answered Jul 30 at 15:22
Lod
286
answered Jul 30 at 15:22
Lod
286
answered Jul 30 at 15:22
Lod
286
286
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
add a comment |Â
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
...which is the same thing, one might add (since $sin(pi-x)=sin x$).
– Hans Lundmark
Jul 30 at 15:42
add a comment |Â
up vote
0
down vote
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $tan$ and $cot$ need to reinforce.
First, some preparation.
We all know that
$$
tan (x pm y) = frac tan (x) pm tan (y) 1 mp tan (x)tan (y).
$$
Then
$$
cot (x - y) = frac cot (x) cot (y) + 1 cot (y) - cot (x),
$$
or
$$
cot (x) cot (y) = cot(x-y) (cot(y) -cot(x)) - 1
$$
Especially,
$$
cot (2x) = frac cot^2 (x) -1 2 cot (x),
$$
which yields
$$
cot^2(x) - 1 = 2cot(x)cot(2x).
$$
Also,
$$
csc^2(x) = frac sin^2(x) + cos^2(x) sin ^2(x) = 1 + cot^2(x).
$$
Now let's start. Let $x = pi /7$.
$$
P := csc^2(x) + csc^2(2x) + csc^2(4x) = 3 + cot^2(x) + cot^2(2x) + cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(cot(x) cot(2x) + cot(2x) cot (4x) + cot (4x) cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $cot(pi pm y) = mpcot (y)$, we have
$$
Q = cot(x)cot(2x) - cot(2x) cot (3x) - cot(3x) cot (x).
$$
Now,
beginalign*
Q &= 1+ cot(x)(cot(x) - cot(2x)) - cot(x) (cot(2x) -cot(3x)) - cot(2x)(cot(x) - cot (3x))\
&= 1+ cot(x) (cot(x) -3cot(2x) + cot(3x)) + cot(2x) cot(3x)\
&= cot(x) (cot(x) -3cot(2x) + cot(3x)) + (cot(2x)-cot(3x))cot(x) \
&= cot(x)(cot(x) -2cot(2x))\
&= cot^2(x) - 2cot(2x) cot(x)\
&=1.
endalign*
Therefore $P = 6+2Q=8$.
answered Jul 30 at 17:19
xbh
1,0257
add a comment |Â
up vote
0
down vote
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $tan$ and $cot$ need to reinforce.
First, some preparation.
We all know that
$$
tan (x pm y) = frac tan (x) pm tan (y) 1 mp tan (x)tan (y).
$$
Then
$$
cot (x - y) = frac cot (x) cot (y) + 1 cot (y) - cot (x),
$$
or
$$
cot (x) cot (y) = cot(x-y) (cot(y) -cot(x)) - 1
$$
Especially,
$$
cot (2x) = frac cot^2 (x) -1 2 cot (x),
$$
which yields
$$
cot^2(x) - 1 = 2cot(x)cot(2x).
$$
Also,
$$
csc^2(x) = frac sin^2(x) + cos^2(x) sin ^2(x) = 1 + cot^2(x).
$$
Now let's start. Let $x = pi /7$.
$$
P := csc^2(x) + csc^2(2x) + csc^2(4x) = 3 + cot^2(x) + cot^2(2x) + cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(cot(x) cot(2x) + cot(2x) cot (4x) + cot (4x) cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $cot(pi pm y) = mpcot (y)$, we have
$$
Q = cot(x)cot(2x) - cot(2x) cot (3x) - cot(3x) cot (x).
$$
Now,
beginalign*
Q &= 1+ cot(x)(cot(x) - cot(2x)) - cot(x) (cot(2x) -cot(3x)) - cot(2x)(cot(x) - cot (3x))\
&= 1+ cot(x) (cot(x) -3cot(2x) + cot(3x)) + cot(2x) cot(3x)\
&= cot(x) (cot(x) -3cot(2x) + cot(3x)) + (cot(2x)-cot(3x))cot(x) \
&= cot(x)(cot(x) -2cot(2x))\
&= cot^2(x) - 2cot(2x) cot(x)\
&=1.
endalign*
Therefore $P = 6+2Q=8$.
answered Jul 30 at 17:19
xbh
1,0257
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $tan$ and $cot$ need to reinforce.
First, some preparation.
We all know that
$$
tan (x pm y) = frac tan (x) pm tan (y) 1 mp tan (x)tan (y).
$$
Then
$$
cot (x - y) = frac cot (x) cot (y) + 1 cot (y) - cot (x),
$$
or
$$
cot (x) cot (y) = cot(x-y) (cot(y) -cot(x)) - 1
$$
Especially,
$$
cot (2x) = frac cot^2 (x) -1 2 cot (x),
$$
which yields
$$
cot^2(x) - 1 = 2cot(x)cot(2x).
$$
Also,
$$
csc^2(x) = frac sin^2(x) + cos^2(x) sin ^2(x) = 1 + cot^2(x).
$$
Now let's start. Let $x = pi /7$.
$$
P := csc^2(x) + csc^2(2x) + csc^2(4x) = 3 + cot^2(x) + cot^2(2x) + cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(cot(x) cot(2x) + cot(2x) cot (4x) + cot (4x) cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $cot(pi pm y) = mpcot (y)$, we have
$$
Q = cot(x)cot(2x) - cot(2x) cot (3x) - cot(3x) cot (x).
$$
Now,
beginalign*
Q &= 1+ cot(x)(cot(x) - cot(2x)) - cot(x) (cot(2x) -cot(3x)) - cot(2x)(cot(x) - cot (3x))\
&= 1+ cot(x) (cot(x) -3cot(2x) + cot(3x)) + cot(2x) cot(3x)\
&= cot(x) (cot(x) -3cot(2x) + cot(3x)) + (cot(2x)-cot(3x))cot(x) \
&= cot(x)(cot(x) -2cot(2x))\
&= cot^2(x) - 2cot(2x) cot(x)\
&=1.
endalign*
Therefore $P = 6+2Q=8$.
answered Jul 30 at 17:19
xbh
1,0257
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $tan$ and $cot$ need to reinforce.
First, some preparation.
We all know that
$$
tan (x pm y) = frac tan (x) pm tan (y) 1 mp tan (x)tan (y).
$$
Then
$$
cot (x - y) = frac cot (x) cot (y) + 1 cot (y) - cot (x),
$$
or
$$
cot (x) cot (y) = cot(x-y) (cot(y) -cot(x)) - 1
$$
Especially,
$$
cot (2x) = frac cot^2 (x) -1 2 cot (x),
$$
which yields
$$
cot^2(x) - 1 = 2cot(x)cot(2x).
$$
Also,
$$
csc^2(x) = frac sin^2(x) + cos^2(x) sin ^2(x) = 1 + cot^2(x).
$$
Now let's start. Let $x = pi /7$.
$$
P := csc^2(x) + csc^2(2x) + csc^2(4x) = 3 + cot^2(x) + cot^2(2x) + cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(cot(x) cot(2x) + cot(2x) cot (4x) + cot (4x) cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $cot(pi pm y) = mpcot (y)$, we have
$$
Q = cot(x)cot(2x) - cot(2x) cot (3x) - cot(3x) cot (x).
$$
Now,
beginalign*
Q &= 1+ cot(x)(cot(x) - cot(2x)) - cot(x) (cot(2x) -cot(3x)) - cot(2x)(cot(x) - cot (3x))\
&= 1+ cot(x) (cot(x) -3cot(2x) + cot(3x)) + cot(2x) cot(3x)\
&= cot(x) (cot(x) -3cot(2x) + cot(3x)) + (cot(2x)-cot(3x))cot(x) \
&= cot(x)(cot(x) -2cot(2x))\
&= cot^2(x) - 2cot(2x) cot(x)\
&=1.
endalign*
Therefore $P = 6+2Q=8$.
answered Jul 30 at 17:19
xbh
1,0257
answered Jul 30 at 17:19
xbh
1,0257
answered Jul 30 at 17:19
xbh
1,0257
answered Jul 30 at 17:19
xbh
1,0257
1,0257
add a comment |Â
add a comment |Â
1
Hi, welcome to MSE. For a proficient interaction, please take a few minutes to learn type MathJax. math.meta.stackexchange.com/questions/5020/…
– xbh
Jul 30 at 15:10
It's not a duplicate just want to know the answer I was confused whether 3pi/7 and 4pi/7 are same or not in case of cosec
– user580093
Jul 30 at 15:45
@user580093 This is the same, 'cause $csc^2(pi - x) = csc^2 (x)$.
– xbh
Jul 30 at 15:57