Proving that two determinants are equal without expanding them
Clash Royale CLAN TAG#URR8PPP
$begingroup$
So I need to prove that
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix
$$
$$
= beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix
$$
Now,
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix = beginvmatrix
sin^2(alpha) & cos^2(alpha) - sin^2(alpha) & cos^2(alpha) \
sin^2(beta) & cos^2(beta) - sin^2(beta) & cos^2(beta) \
sin^2(gamma) & cos^2(gamma) - sin^2(gamma) & cos^2(gamma) \
endvmatrix
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
linear-algebra determinant
asked Feb 16 at 11:39
HelixHelix
1558
$endgroup$
add a comment |
$begingroup$
So I need to prove that
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix
$$
$$
= beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix
$$
Now,
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix = beginvmatrix
sin^2(alpha) & cos^2(alpha) - sin^2(alpha) & cos^2(alpha) \
sin^2(beta) & cos^2(beta) - sin^2(beta) & cos^2(beta) \
sin^2(gamma) & cos^2(gamma) - sin^2(gamma) & cos^2(gamma) \
endvmatrix
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
linear-algebra determinant
asked Feb 16 at 11:39
HelixHelix
1558
$endgroup$
2
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47
add a comment |
$begingroup$
So I need to prove that
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix
$$
$$
= beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix
$$
Now,
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix = beginvmatrix
sin^2(alpha) & cos^2(alpha) - sin^2(alpha) & cos^2(alpha) \
sin^2(beta) & cos^2(beta) - sin^2(beta) & cos^2(beta) \
sin^2(gamma) & cos^2(gamma) - sin^2(gamma) & cos^2(gamma) \
endvmatrix
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
linear-algebra determinant
asked Feb 16 at 11:39
HelixHelix
1558
$endgroup$
So I need to prove that
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix
$$
$$
= beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix
$$
Now,
$$
beginvmatrix
sin^2(alpha) & cos(2alpha) & cos^2(alpha) \
sin^2(beta) & cos(2beta) & cos^2(beta) \
sin^2(gamma) & cos(2gamma) & cos^2(gamma) \
endvmatrix = beginvmatrix
sin^2(alpha) & cos^2(alpha) - sin^2(alpha) & cos^2(alpha) \
sin^2(beta) & cos^2(beta) - sin^2(beta) & cos^2(beta) \
sin^2(gamma) & cos^2(gamma) - sin^2(gamma) & cos^2(gamma) \
endvmatrix
$$
Adding column $1$ to column $2$ then makes column $2$ and column $3$ equal and hence the first determinant $ = 0$.
I'm stuck in trying to prove that the second determinant is also zero, so I need help in that.
linear-algebra determinant
linear-algebra determinant
asked Feb 16 at 11:39
HelixHelix
1558
asked Feb 16 at 11:39
HelixHelix
1558
asked Feb 16 at 11:39
HelixHelix
1558
asked Feb 16 at 11:39
HelixHelix
1558
asked Feb 16 at 11:39
HelixHelix
1558
1558
2
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47
add a comment |
2
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47
2
2
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix=beginvmatrix
sin(alpha) & cos(alpha) & sinalphacolorredcosdelta+colorredsindeltacosalpha \
sin(beta) & cos(beta) & sinbetacolorredcosdelta+colorredsindeltacosbeta \
sin(gamma) & cos(gamma) & singammacolorredcosdelta+colorredsindeltacosgamma \
endvmatrix$$$$$$
$$=beginvmatrix
sin(alpha) & cos(alpha) & sinalpha \
sin(beta) & cos(beta) & sinbeta\
sin(gamma) & cos(gamma) & singamma \
endvmatrixcolorredcosdelta+beginvmatrix
sin(alpha) & cos(alpha) & cosalpha \
sin(beta) & cos(beta) & cosbeta\
sin(gamma) & cos(gamma) & cosgamma \
endvmatrixcolorredsindelta=0+0=0$$
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
|
show 4 more comments
$begingroup$
HINT: Relate the third column to the first two using the trigonometric identity
$$sin(theta+delta)=sinthetacosdelta+costhetasindelta.$$
$endgroup$
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3114955%2fproving-that-two-determinants-are-equal-without-expanding-them%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix=beginvmatrix
sin(alpha) & cos(alpha) & sinalphacolorredcosdelta+colorredsindeltacosalpha \
sin(beta) & cos(beta) & sinbetacolorredcosdelta+colorredsindeltacosbeta \
sin(gamma) & cos(gamma) & singammacolorredcosdelta+colorredsindeltacosgamma \
endvmatrix$$$$$$
$$=beginvmatrix
sin(alpha) & cos(alpha) & sinalpha \
sin(beta) & cos(beta) & sinbeta\
sin(gamma) & cos(gamma) & singamma \
endvmatrixcolorredcosdelta+beginvmatrix
sin(alpha) & cos(alpha) & cosalpha \
sin(beta) & cos(beta) & cosbeta\
sin(gamma) & cos(gamma) & cosgamma \
endvmatrixcolorredsindelta=0+0=0$$
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
|
show 4 more comments
$begingroup$
$$beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix=beginvmatrix
sin(alpha) & cos(alpha) & sinalphacolorredcosdelta+colorredsindeltacosalpha \
sin(beta) & cos(beta) & sinbetacolorredcosdelta+colorredsindeltacosbeta \
sin(gamma) & cos(gamma) & singammacolorredcosdelta+colorredsindeltacosgamma \
endvmatrix$$$$$$
$$=beginvmatrix
sin(alpha) & cos(alpha) & sinalpha \
sin(beta) & cos(beta) & sinbeta\
sin(gamma) & cos(gamma) & singamma \
endvmatrixcolorredcosdelta+beginvmatrix
sin(alpha) & cos(alpha) & cosalpha \
sin(beta) & cos(beta) & cosbeta\
sin(gamma) & cos(gamma) & cosgamma \
endvmatrixcolorredsindelta=0+0=0$$
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
|
show 4 more comments
$begingroup$
$$beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix=beginvmatrix
sin(alpha) & cos(alpha) & sinalphacolorredcosdelta+colorredsindeltacosalpha \
sin(beta) & cos(beta) & sinbetacolorredcosdelta+colorredsindeltacosbeta \
sin(gamma) & cos(gamma) & singammacolorredcosdelta+colorredsindeltacosgamma \
endvmatrix$$$$$$
$$=beginvmatrix
sin(alpha) & cos(alpha) & sinalpha \
sin(beta) & cos(beta) & sinbeta\
sin(gamma) & cos(gamma) & singamma \
endvmatrixcolorredcosdelta+beginvmatrix
sin(alpha) & cos(alpha) & cosalpha \
sin(beta) & cos(beta) & cosbeta\
sin(gamma) & cos(gamma) & cosgamma \
endvmatrixcolorredsindelta=0+0=0$$
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
$endgroup$
$$beginvmatrix
sin(alpha) & cos(alpha) & sin(alpha + delta) \
sin(beta) & cos(beta) & sin(beta + delta) \
sin(gamma) & cos(gamma) & sin(gamma + delta) \
endvmatrix=beginvmatrix
sin(alpha) & cos(alpha) & sinalphacolorredcosdelta+colorredsindeltacosalpha \
sin(beta) & cos(beta) & sinbetacolorredcosdelta+colorredsindeltacosbeta \
sin(gamma) & cos(gamma) & singammacolorredcosdelta+colorredsindeltacosgamma \
endvmatrix$$$$$$
$$=beginvmatrix
sin(alpha) & cos(alpha) & sinalpha \
sin(beta) & cos(beta) & sinbeta\
sin(gamma) & cos(gamma) & singamma \
endvmatrixcolorredcosdelta+beginvmatrix
sin(alpha) & cos(alpha) & cosalpha \
sin(beta) & cos(beta) & cosbeta\
sin(gamma) & cos(gamma) & cosgamma \
endvmatrixcolorredsindelta=0+0=0$$
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
edited Feb 16 at 11:53
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
answered Feb 16 at 11:50
DonAntonioDonAntonio
179k1494233
179k1494233
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
|
show 4 more comments
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
I completely forgot about that property, thank you.
$endgroup$
– Helix
Feb 16 at 11:52
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
...but that's precisely the property you used to prove that the first determinant equals $0$...
$endgroup$
– Servaes
Feb 16 at 11:53
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
$begingroup$
Servaes is right. More than forget that property I think all those sines and cosines together confused you.
$endgroup$
– DonAntonio
Feb 16 at 11:54
1
1
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@Servaes I meant the one where you can make two determinants from one by splitting the elements of a column or row
$endgroup$
– Helix
Feb 16 at 11:54
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
$begingroup$
@DonAntonio That's probably it, thanks anyways.
$endgroup$
– Helix
Feb 16 at 11:56
|
show 4 more comments
$begingroup$
HINT: Relate the third column to the first two using the trigonometric identity
$$sin(theta+delta)=sinthetacosdelta+costhetasindelta.$$
$endgroup$
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
add a comment |
$begingroup$
HINT: Relate the third column to the first two using the trigonometric identity
$$sin(theta+delta)=sinthetacosdelta+costhetasindelta.$$
$endgroup$
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
add a comment |
$begingroup$
HINT: Relate the third column to the first two using the trigonometric identity
$$sin(theta+delta)=sinthetacosdelta+costhetasindelta.$$
$endgroup$
HINT: Relate the third column to the first two using the trigonometric identity
$$sin(theta+delta)=sinthetacosdelta+costhetasindelta.$$
answered Feb 16 at 11:45
community wiki
Servaes
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
add a comment |
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
That's what I said, innit?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
$begingroup$
@GerryMyerson I guess it pretty much is. Would you like the give the answer? I can delete this one.
$endgroup$
– Servaes
Feb 16 at 11:48
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3114955%2fproving-that-two-determinants-are-equal-without-expanding-them%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
Expand $sin(alpha+delta)$, note relation to first two columns.
$endgroup$
– Gerry Myerson
Feb 16 at 11:42
$begingroup$
@Gerry I tried doing that, however it didn't lead me anywhere. What do I do after that?
$endgroup$
– Helix
Feb 16 at 11:43
$begingroup$
What did you get when you expanded $sin(alpha+delta)$?
$endgroup$
– Gerry Myerson
Feb 16 at 11:46
$begingroup$
$sin(alpha)cos(delta) + sin(delta)cos(alpha)$. Still kind of confused as to how to use the relation
$endgroup$
– Helix
Feb 16 at 11:47