Mixing
How to prove that $cot{frac{11pi}{18}} +cot{frac{2pi}{9}}=4sin{frac{pi}{18}}cot{frac{2pi}{9}}$
$begingroup$
How to prove this trigonometric identities ?
$$cot{frac{11pi}{18}} +cot{frac{2pi}{9}}=4sin{frac{pi}{18}}cot{frac{2pi}{9}}$$
Thank you.
trigonometry
asked Feb 13 '15 at 10:19
kongkong
27528
$endgroup$
add a comment |
$begingroup$
How to prove this trigonometric identities ?
$$cot{frac{11pi}{18}} +cot{frac{2pi}{9}}=4sin{frac{pi}{18}}cot{frac{2pi}{9}}$$
Thank you.
trigonometry
asked Feb 13 '15 at 10:19
kongkong
27528
$endgroup$
$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26
add a comment |
$begingroup$
How to prove this trigonometric identities ?
$$cot{frac{11pi}{18}} +cot{frac{2pi}{9}}=4sin{frac{pi}{18}}cot{frac{2pi}{9}}$$
Thank you.
trigonometry
asked Feb 13 '15 at 10:19
kongkong
27528
$endgroup$
How to prove this trigonometric identities ?
$$cot{frac{11pi}{18}} +cot{frac{2pi}{9}}=4sin{frac{pi}{18}}cot{frac{2pi}{9}}$$
Thank you.
trigonometry
trigonometry
asked Feb 13 '15 at 10:19
kongkong
27528
asked Feb 13 '15 at 10:19
kongkong
27528
asked Feb 13 '15 at 10:19
kongkong
27528
asked Feb 13 '15 at 10:19
kongkong
27528
asked Feb 13 '15 at 10:19
kongkong
27528
27528
$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26
add a comment |
$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26
$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26
$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have to check:
$$ -tanfrac{pi}{9}+cotfrac{2pi}{9}=4cosfrac{4pi}{9}cotfrac{2pi}{9} $$
hence, by multiplying both sides by $sinfrac{2pi}{9}=2sinfrac{pi}{9}cosfrac{pi}{9}$:
$$ -2sin^2frac{pi}{9}+cosfrac{2pi}{9} = 4cosfrac{4pi}{9}cosfrac{2pi}{9} $$
so we just have to check that $x=cosfrac{2pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2cdot T_3(x)-1, tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $cosfrac{2pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(cosalpha)=cos(3alpha)$:
$$ 2cdot T_3left(cosfrac{2pi}{9}right)-1 = 2cdotcosfrac{2pi}{3}-1 = 0, $$
hence we're done.
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
add a comment |
$begingroup$
$cotdfrac{11pi}{18}=cotleft(pi-dfrac{7pi}{18}right)=-cotdfrac{7pi}{18}$
Now $cotdfrac{2pi}9-cotdfrac{7pi}{18}=dfrac{sinleft(dfrac{7pi}{18}-dfrac{2pi}9right)}{sindfrac{2pi}9cdotsindfrac{7pi}{18}}=dfrac1{2sindfrac{2pi}9cosdfracpi9}$
( as $sindfrac{7pi}{18}=cosleft(dfracpi2-dfrac{7pi}{18}right)=cosdfracpi9$)
which needs to be $=sindfracpi{18}cotdfrac{2pi}9$
which will be true if $8sindfracpi{18}cosdfracpi9cosdfrac{2pi}9=1$
$iff8cosdfracpi9cosdfrac{2pi}9cosdfrac{4pi}9=1$
as $dfrac{4pi}9+dfracpi{18}=dfracpi2$
Now use: Upon multiplying $cos(20^circ)cos(40^circ)cos(80^circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
edited Apr 13 '17 at 12:20
Community♦
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have to check:
$$ -tanfrac{pi}{9}+cotfrac{2pi}{9}=4cosfrac{4pi}{9}cotfrac{2pi}{9} $$
hence, by multiplying both sides by $sinfrac{2pi}{9}=2sinfrac{pi}{9}cosfrac{pi}{9}$:
$$ -2sin^2frac{pi}{9}+cosfrac{2pi}{9} = 4cosfrac{4pi}{9}cosfrac{2pi}{9} $$
so we just have to check that $x=cosfrac{2pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2cdot T_3(x)-1, tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $cosfrac{2pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(cosalpha)=cos(3alpha)$:
$$ 2cdot T_3left(cosfrac{2pi}{9}right)-1 = 2cdotcosfrac{2pi}{3}-1 = 0, $$
hence we're done.
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
add a comment |
$begingroup$
We have to check:
$$ -tanfrac{pi}{9}+cotfrac{2pi}{9}=4cosfrac{4pi}{9}cotfrac{2pi}{9} $$
hence, by multiplying both sides by $sinfrac{2pi}{9}=2sinfrac{pi}{9}cosfrac{pi}{9}$:
$$ -2sin^2frac{pi}{9}+cosfrac{2pi}{9} = 4cosfrac{4pi}{9}cosfrac{2pi}{9} $$
so we just have to check that $x=cosfrac{2pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2cdot T_3(x)-1, tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $cosfrac{2pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(cosalpha)=cos(3alpha)$:
$$ 2cdot T_3left(cosfrac{2pi}{9}right)-1 = 2cdotcosfrac{2pi}{3}-1 = 0, $$
hence we're done.
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
add a comment |
$begingroup$
We have to check:
$$ -tanfrac{pi}{9}+cotfrac{2pi}{9}=4cosfrac{4pi}{9}cotfrac{2pi}{9} $$
hence, by multiplying both sides by $sinfrac{2pi}{9}=2sinfrac{pi}{9}cosfrac{pi}{9}$:
$$ -2sin^2frac{pi}{9}+cosfrac{2pi}{9} = 4cosfrac{4pi}{9}cosfrac{2pi}{9} $$
so we just have to check that $x=cosfrac{2pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2cdot T_3(x)-1, tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $cosfrac{2pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(cosalpha)=cos(3alpha)$:
$$ 2cdot T_3left(cosfrac{2pi}{9}right)-1 = 2cdotcosfrac{2pi}{3}-1 = 0, $$
hence we're done.
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
We have to check:
$$ -tanfrac{pi}{9}+cotfrac{2pi}{9}=4cosfrac{4pi}{9}cotfrac{2pi}{9} $$
hence, by multiplying both sides by $sinfrac{2pi}{9}=2sinfrac{pi}{9}cosfrac{pi}{9}$:
$$ -2sin^2frac{pi}{9}+cosfrac{2pi}{9} = 4cosfrac{4pi}{9}cosfrac{2pi}{9} $$
so we just have to check that $x=cosfrac{2pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2cdot T_3(x)-1, tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $cosfrac{2pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(cosalpha)=cos(3alpha)$:
$$ 2cdot T_3left(cosfrac{2pi}{9}right)-1 = 2cdotcosfrac{2pi}{3}-1 = 0, $$
hence we're done.
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
answered Feb 13 '15 at 10:45
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
add a comment |
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
Thank you. Are there pure trigonometric methods or another?
$endgroup$
– kong
Feb 13 '15 at 11:28
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
$begingroup$
@kong, Please find the other solution.
$endgroup$
– lab bhattacharjee
Feb 14 '15 at 16:35
add a comment |
$begingroup$
$cotdfrac{11pi}{18}=cotleft(pi-dfrac{7pi}{18}right)=-cotdfrac{7pi}{18}$
Now $cotdfrac{2pi}9-cotdfrac{7pi}{18}=dfrac{sinleft(dfrac{7pi}{18}-dfrac{2pi}9right)}{sindfrac{2pi}9cdotsindfrac{7pi}{18}}=dfrac1{2sindfrac{2pi}9cosdfracpi9}$
( as $sindfrac{7pi}{18}=cosleft(dfracpi2-dfrac{7pi}{18}right)=cosdfracpi9$)
which needs to be $=sindfracpi{18}cotdfrac{2pi}9$
which will be true if $8sindfracpi{18}cosdfracpi9cosdfrac{2pi}9=1$
$iff8cosdfracpi9cosdfrac{2pi}9cosdfrac{4pi}9=1$
as $dfrac{4pi}9+dfracpi{18}=dfracpi2$
Now use: Upon multiplying $cos(20^circ)cos(40^circ)cos(80^circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
edited Apr 13 '17 at 12:20
Community♦
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
add a comment |
$begingroup$
$cotdfrac{11pi}{18}=cotleft(pi-dfrac{7pi}{18}right)=-cotdfrac{7pi}{18}$
Now $cotdfrac{2pi}9-cotdfrac{7pi}{18}=dfrac{sinleft(dfrac{7pi}{18}-dfrac{2pi}9right)}{sindfrac{2pi}9cdotsindfrac{7pi}{18}}=dfrac1{2sindfrac{2pi}9cosdfracpi9}$
( as $sindfrac{7pi}{18}=cosleft(dfracpi2-dfrac{7pi}{18}right)=cosdfracpi9$)
which needs to be $=sindfracpi{18}cotdfrac{2pi}9$
which will be true if $8sindfracpi{18}cosdfracpi9cosdfrac{2pi}9=1$
$iff8cosdfracpi9cosdfrac{2pi}9cosdfrac{4pi}9=1$
as $dfrac{4pi}9+dfracpi{18}=dfracpi2$
Now use: Upon multiplying $cos(20^circ)cos(40^circ)cos(80^circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
edited Apr 13 '17 at 12:20
Community♦
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
add a comment |
$begingroup$
$cotdfrac{11pi}{18}=cotleft(pi-dfrac{7pi}{18}right)=-cotdfrac{7pi}{18}$
Now $cotdfrac{2pi}9-cotdfrac{7pi}{18}=dfrac{sinleft(dfrac{7pi}{18}-dfrac{2pi}9right)}{sindfrac{2pi}9cdotsindfrac{7pi}{18}}=dfrac1{2sindfrac{2pi}9cosdfracpi9}$
( as $sindfrac{7pi}{18}=cosleft(dfracpi2-dfrac{7pi}{18}right)=cosdfracpi9$)
which needs to be $=sindfracpi{18}cotdfrac{2pi}9$
which will be true if $8sindfracpi{18}cosdfracpi9cosdfrac{2pi}9=1$
$iff8cosdfracpi9cosdfrac{2pi}9cosdfrac{4pi}9=1$
as $dfrac{4pi}9+dfracpi{18}=dfracpi2$
Now use: Upon multiplying $cos(20^circ)cos(40^circ)cos(80^circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
edited Apr 13 '17 at 12:20
Community♦
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
$endgroup$
$cotdfrac{11pi}{18}=cotleft(pi-dfrac{7pi}{18}right)=-cotdfrac{7pi}{18}$
Now $cotdfrac{2pi}9-cotdfrac{7pi}{18}=dfrac{sinleft(dfrac{7pi}{18}-dfrac{2pi}9right)}{sindfrac{2pi}9cdotsindfrac{7pi}{18}}=dfrac1{2sindfrac{2pi}9cosdfracpi9}$
( as $sindfrac{7pi}{18}=cosleft(dfracpi2-dfrac{7pi}{18}right)=cosdfracpi9$)
which needs to be $=sindfracpi{18}cotdfrac{2pi}9$
which will be true if $8sindfracpi{18}cosdfracpi9cosdfrac{2pi}9=1$
$iff8cosdfracpi9cosdfrac{2pi}9cosdfrac{4pi}9=1$
as $dfrac{4pi}9+dfracpi{18}=dfracpi2$
Now use: Upon multiplying $cos(20^circ)cos(40^circ)cos(80^circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
edited Apr 13 '17 at 12:20
Community♦
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
answered Feb 14 '15 at 16:34
lab bhattacharjeelab bhattacharjee
228k15158279
228k15158279
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
add a comment |
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
$begingroup$
@kong, How about this?
$endgroup$
– lab bhattacharjee
Feb 21 '15 at 5:03
add a comment |
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$begingroup$
Have you tried anything? I believe you should take $phi = pi/18$ for simplicity, write the problem in terms of $phi$ and then just use simple formulas like $cot phi = cos phi/sin phi$, $cos alpha sin beta = sin frac{alpha + beta}{2}sin frac{alpha-beta}{2}$ and so on.
$endgroup$
– Andrei Rykhalski
Feb 13 '15 at 10:26