Proving determinants is independent of thetaAn unusual symmetric inequality of trigonometric functionsA determinant problem$2sin(theta + 17) = dfrac cos (theta +8)cos (theta + 17)$Given Triangle $a,b,c$ and angles $theta,alpha,beta$.Find $fraccos(alpha)sin(theta)$Prove that $alpha=beta=gamma$How to proof this trig equality???(question 1 : solved)(question 2 : solved)Proving the following identity$min$ of expression $sin alpha+sin beta+sin gamma,$satisfying $alpha+beta+gamma = pi$Sum this series: $1+costhetasectheta+cos2thetasec^2theta+cdots+cosnthetasec^ntheta$Show the following matrix has determinant = 0
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Proving determinants is independent of theta
An unusual symmetric inequality of trigonometric functionsA determinant problem$2sin(theta + 17) = dfrac cos (theta +8)cos (theta + 17)$Given Triangle $a,b,c$ and angles $theta,alpha,beta$.Find $fraccos(alpha)sin(theta)$Prove that $alpha=beta=gamma$How to proof this trig equality???(question 1 : solved)(question 2 : solved)Proving the following identity$min$ of expression $sin alpha+sin beta+sin gamma,$satisfying $alpha+beta+gamma = pi$Sum this series: $1+costhetasectheta+cos2thetasec^2theta+cdots+cosnthetasec^ntheta$Show the following matrix has determinant = 0
$begingroup$
$$A=left|beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright|$$
My approach is to subtract row 1 from row 2 and row 3 then open the determinant. The problem with this approach is it involves way too much calculation. I tried to take something common but nothing comes to mind. Can somebody give me a better way to approach this problem?
trigonometry determinant
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
$$A=left|beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright|$$
My approach is to subtract row 1 from row 2 and row 3 then open the determinant. The problem with this approach is it involves way too much calculation. I tried to take something common but nothing comes to mind. Can somebody give me a better way to approach this problem?
trigonometry determinant
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
4
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
$endgroup$
– Blue
May 12 at 19:57
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
$endgroup$
– B Luthra
May 12 at 20:03
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04
add a comment |
$begingroup$
$$A=left|beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright|$$
My approach is to subtract row 1 from row 2 and row 3 then open the determinant. The problem with this approach is it involves way too much calculation. I tried to take something common but nothing comes to mind. Can somebody give me a better way to approach this problem?
trigonometry determinant
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$$A=left|beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright|$$
My approach is to subtract row 1 from row 2 and row 3 then open the determinant. The problem with this approach is it involves way too much calculation. I tried to take something common but nothing comes to mind. Can somebody give me a better way to approach this problem?
trigonometry determinant
trigonometry determinant
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
edited May 12 at 19:44
Jean-Claude Arbaut
15.8k63865
15.8k63865
asked May 12 at 19:19
B LuthraB Luthra
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked May 12 at 19:19
B LuthraB Luthra
183
asked May 12 at 19:19
B LuthraB Luthra
183
183
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
B Luthra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
4
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
$endgroup$
– Blue
May 12 at 19:57
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
$endgroup$
– B Luthra
May 12 at 20:03
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04
add a comment |
$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
4
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
$endgroup$
– Blue
May 12 at 19:57
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
$endgroup$
– B Luthra
May 12 at 20:03
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04
$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
4
4
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
$endgroup$
– Blue
May 12 at 19:57
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
$endgroup$
– Blue
May 12 at 19:57
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
$endgroup$
– B Luthra
May 12 at 20:03
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
$endgroup$
– B Luthra
May 12 at 20:03
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.
If you want to make this rigorous, observe that
$$left(beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright) = left(beginarrayccc
sinleft(alpharight) & cosleft(alpharight) & 1\
sinleft(betaright) & cosleft(betaright) & 1\
sinleft(gammaright) & cosleft(gammaright) & 1
endarrayright)left(beginarrayccc
cosleft(thetaright) & -sinleft(thetaright) & 0\
sinleft(thetaright) & cosleft(thetaright) & 0\
0 & 0 & 1
endarrayright)$$
and use $det(AB) = det(A) det (B)$.
(See also Blue's comment above that beat me to the punch.)
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
$endgroup$
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
|
show 3 more comments
$begingroup$
Expand along the last column (using Laplace's formula).
One minor is for instance
$$sin(theta+beta)cos(theta+gamma)-sin(theta+gamma)cos(theta+beta)\=
sin[(theta+beta)-(theta+gamma)]\=
sin(beta-gamma)$$
The other two minors are similar. All in all you get:
$$|A|=sin(beta-gamma)-sin(alpha-gamma)+sin(alpha-beta)$$
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
$endgroup$
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
add a comment |
$begingroup$
Idea: If $A =det (c_1,c_2,cdots,c_n)$, where $c_k$ are columns (of functions with $theta$ as argument), then a general fact is that (product/sum rules) $$A' = det (c_1',c_2,cdots,c_n)+ det (c_1, c_2',cdots, c_n) + cdots + det (c_1, c_2,cdots, c_n').$$
In the case here, each of the summands will be visibly zero. Hence $A'(theta) = 0$, and $A$ does not depend on $theta$.
answered May 12 at 20:19
peter a gpeter a g
3,3151714
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.
If you want to make this rigorous, observe that
$$left(beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright) = left(beginarrayccc
sinleft(alpharight) & cosleft(alpharight) & 1\
sinleft(betaright) & cosleft(betaright) & 1\
sinleft(gammaright) & cosleft(gammaright) & 1
endarrayright)left(beginarrayccc
cosleft(thetaright) & -sinleft(thetaright) & 0\
sinleft(thetaright) & cosleft(thetaright) & 0\
0 & 0 & 1
endarrayright)$$
and use $det(AB) = det(A) det (B)$.
(See also Blue's comment above that beat me to the punch.)
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
$endgroup$
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
|
show 3 more comments
$begingroup$
A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.
If you want to make this rigorous, observe that
$$left(beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright) = left(beginarrayccc
sinleft(alpharight) & cosleft(alpharight) & 1\
sinleft(betaright) & cosleft(betaright) & 1\
sinleft(gammaright) & cosleft(gammaright) & 1
endarrayright)left(beginarrayccc
cosleft(thetaright) & -sinleft(thetaright) & 0\
sinleft(thetaright) & cosleft(thetaright) & 0\
0 & 0 & 1
endarrayright)$$
and use $det(AB) = det(A) det (B)$.
(See also Blue's comment above that beat me to the punch.)
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
$endgroup$
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
|
show 3 more comments
$begingroup$
A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.
If you want to make this rigorous, observe that
$$left(beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright) = left(beginarrayccc
sinleft(alpharight) & cosleft(alpharight) & 1\
sinleft(betaright) & cosleft(betaright) & 1\
sinleft(gammaright) & cosleft(gammaright) & 1
endarrayright)left(beginarrayccc
cosleft(thetaright) & -sinleft(thetaright) & 0\
sinleft(thetaright) & cosleft(thetaright) & 0\
0 & 0 & 1
endarrayright)$$
and use $det(AB) = det(A) det (B)$.
(See also Blue's comment above that beat me to the punch.)
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
$endgroup$
A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.
If you want to make this rigorous, observe that
$$left(beginarrayccc
sinleft(theta+alpharight) & cosleft(theta+alpharight) & 1\
sinleft(theta+betaright) & cosleft(theta+betaright) & 1\
sinleft(theta+gammaright) & cosleft(theta+gammaright) & 1
endarrayright) = left(beginarrayccc
sinleft(alpharight) & cosleft(alpharight) & 1\
sinleft(betaright) & cosleft(betaright) & 1\
sinleft(gammaright) & cosleft(gammaright) & 1
endarrayright)left(beginarrayccc
cosleft(thetaright) & -sinleft(thetaright) & 0\
sinleft(thetaright) & cosleft(thetaright) & 0\
0 & 0 & 1
endarrayright)$$
and use $det(AB) = det(A) det (B)$.
(See also Blue's comment above that beat me to the punch.)
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
answered May 12 at 20:19
Jair TaylorJair Taylor
9,30932244
9,30932244
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
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– Jair Taylor
May 13 at 2:03
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this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
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– B Luthra
May 13 at 11:02
|
show 3 more comments
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-sin(theta) should be a21 not a12 rest is fine
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– B Luthra
May 12 at 20:47
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Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
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– Jean-Claude Arbaut
May 12 at 20:51
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@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
-sin(theta) should be a21 not a12 rest is fine
$endgroup$
– B Luthra
May 12 at 20:47
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
Interesting approach. One minor remark: "vectors on the unit circle" is not entirely clear, though in the context I understand what you mean (here there are both the affine space $Bbb R^3$ with a plane $z=1$, and the vector space $Bbb R^3$ with the row vectors of this $3times3$ matrix, but of course there is a relationship between the two spaces).
$endgroup$
– Jean-Claude Arbaut
May 12 at 20:51
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@BLuthra It looks correct to me.
$endgroup$
– Jair Taylor
May 13 at 2:00
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
@Jean-ClaudeArbaut Sure, perhaps "points on the unit circle on the plane $z = 1$" would more clear, since as vectors they are not parallel to the $z=1$ plane.
$endgroup$
– Jair Taylor
May 13 at 2:03
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
$begingroup$
this is a determinant so we will multiply row by row multiply sin(x)*cos(y)-cosx*siny if alpha =x and theta = y will give sin(x-y)but we want sin(x+y)
$endgroup$
– B Luthra
May 13 at 11:02
|
show 3 more comments
$begingroup$
Expand along the last column (using Laplace's formula).
One minor is for instance
$$sin(theta+beta)cos(theta+gamma)-sin(theta+gamma)cos(theta+beta)\=
sin[(theta+beta)-(theta+gamma)]\=
sin(beta-gamma)$$
The other two minors are similar. All in all you get:
$$|A|=sin(beta-gamma)-sin(alpha-gamma)+sin(alpha-beta)$$
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
$endgroup$
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
add a comment |
$begingroup$
Expand along the last column (using Laplace's formula).
One minor is for instance
$$sin(theta+beta)cos(theta+gamma)-sin(theta+gamma)cos(theta+beta)\=
sin[(theta+beta)-(theta+gamma)]\=
sin(beta-gamma)$$
The other two minors are similar. All in all you get:
$$|A|=sin(beta-gamma)-sin(alpha-gamma)+sin(alpha-beta)$$
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
$endgroup$
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
add a comment |
$begingroup$
Expand along the last column (using Laplace's formula).
One minor is for instance
$$sin(theta+beta)cos(theta+gamma)-sin(theta+gamma)cos(theta+beta)\=
sin[(theta+beta)-(theta+gamma)]\=
sin(beta-gamma)$$
The other two minors are similar. All in all you get:
$$|A|=sin(beta-gamma)-sin(alpha-gamma)+sin(alpha-beta)$$
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
$endgroup$
Expand along the last column (using Laplace's formula).
One minor is for instance
$$sin(theta+beta)cos(theta+gamma)-sin(theta+gamma)cos(theta+beta)\=
sin[(theta+beta)-(theta+gamma)]\=
sin(beta-gamma)$$
The other two minors are similar. All in all you get:
$$|A|=sin(beta-gamma)-sin(alpha-gamma)+sin(alpha-beta)$$
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
edited May 12 at 19:35
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
answered May 12 at 19:26
Jean-Claude ArbautJean-Claude Arbaut
15.8k63865
15.8k63865
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
add a comment |
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
What's the the advantage? Nothing get cancel
$endgroup$
– B Luthra
May 12 at 19:31
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
$begingroup$
@BLuthra $theta$ gets canceled, with almost no calculation (you just have to know the formula for $sin(a-b)$). But that's right, you could have found that by your method, or even by Sarrus' rule, however with some more computations. Here the minors are pretty obvious.
$endgroup$
– Jean-Claude Arbaut
May 12 at 19:37
add a comment |
$begingroup$
Idea: If $A =det (c_1,c_2,cdots,c_n)$, where $c_k$ are columns (of functions with $theta$ as argument), then a general fact is that (product/sum rules) $$A' = det (c_1',c_2,cdots,c_n)+ det (c_1, c_2',cdots, c_n) + cdots + det (c_1, c_2,cdots, c_n').$$
In the case here, each of the summands will be visibly zero. Hence $A'(theta) = 0$, and $A$ does not depend on $theta$.
answered May 12 at 20:19
peter a gpeter a g
3,3151714
$endgroup$
add a comment |
$begingroup$
Idea: If $A =det (c_1,c_2,cdots,c_n)$, where $c_k$ are columns (of functions with $theta$ as argument), then a general fact is that (product/sum rules) $$A' = det (c_1',c_2,cdots,c_n)+ det (c_1, c_2',cdots, c_n) + cdots + det (c_1, c_2,cdots, c_n').$$
In the case here, each of the summands will be visibly zero. Hence $A'(theta) = 0$, and $A$ does not depend on $theta$.
answered May 12 at 20:19
peter a gpeter a g
3,3151714
$endgroup$
add a comment |
$begingroup$
Idea: If $A =det (c_1,c_2,cdots,c_n)$, where $c_k$ are columns (of functions with $theta$ as argument), then a general fact is that (product/sum rules) $$A' = det (c_1',c_2,cdots,c_n)+ det (c_1, c_2',cdots, c_n) + cdots + det (c_1, c_2,cdots, c_n').$$
In the case here, each of the summands will be visibly zero. Hence $A'(theta) = 0$, and $A$ does not depend on $theta$.
answered May 12 at 20:19
peter a gpeter a g
3,3151714
$endgroup$
Idea: If $A =det (c_1,c_2,cdots,c_n)$, where $c_k$ are columns (of functions with $theta$ as argument), then a general fact is that (product/sum rules) $$A' = det (c_1',c_2,cdots,c_n)+ det (c_1, c_2',cdots, c_n) + cdots + det (c_1, c_2,cdots, c_n').$$
In the case here, each of the summands will be visibly zero. Hence $A'(theta) = 0$, and $A$ does not depend on $theta$.
answered May 12 at 20:19
peter a gpeter a g
3,3151714
edited May 13 at 13:10
answered May 12 at 20:19
peter a gpeter a g
3,3151714
answered May 12 at 20:19
peter a gpeter a g
3,3151714
answered May 12 at 20:19
peter a gpeter a g
3,3151714
3,3151714
add a comment |
add a comment |
B Luthra is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I've rewritten it in MathJax.
$endgroup$
– J.G.
May 12 at 19:25
$begingroup$
@BLuthra Are you allowed / do you know calculus? The derivative of $A$ wrt $theta$ is visibly zero
$endgroup$
– peter a g
May 12 at 19:52
4
$begingroup$
The determinant gives (up to a scalar multiple) the volume of a tetrahedron with vertices $(cos(alpha+theta), sin(alpha+theta), 1)$, etc, and the origin. Rotating the tetrahedron through angle $theta$ about the $z$ axis moves the non-origin vertices to $(cosalpha, sinalpha, 1)$, etc. Clearly, the volume doesn't change, but the determinant is freed of $theta$.
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– Blue
May 12 at 19:57
$begingroup$
@peterag we have to prove determinant is independent of theta not equal to zero
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– B Luthra
May 12 at 20:03
$begingroup$
Yes, if the derivative of $A$ is zero, then the function $A$ is constant.
$endgroup$
– peter a g
May 12 at 20:04