Given $cos^2 x = 2 sin x cos x$, why can't I cancel $cos x$ to get $cos x = 2 sin x$?
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If I have a function where I know $cos^2 x = 2 sin x cos x$. Why can I not cross out $cos x$ on both sides, because I get different values for $cos x = 2 sin x$?
algebra-precalculus trigonometry
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
 |Â
show 1 more comment
up vote
3
down vote
favorite
If I have a function where I know $cos^2 x = 2 sin x cos x$. Why can I not cross out $cos x$ on both sides, because I get different values for $cos x = 2 sin x$?
algebra-precalculus trigonometry
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
1
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If I have a function where I know $cos^2 x = 2 sin x cos x$. Why can I not cross out $cos x$ on both sides, because I get different values for $cos x = 2 sin x$?
algebra-precalculus trigonometry
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
If I have a function where I know $cos^2 x = 2 sin x cos x$. Why can I not cross out $cos x$ on both sides, because I get different values for $cos x = 2 sin x$?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
edited Sep 7 at 4:41
Jyrki Lahtonen
106k12163358
106k12163358
asked Sep 7 at 0:52
Juliana Guerra
182
asked Sep 7 at 0:52
Juliana Guerra
182
asked Sep 7 at 0:52
Juliana Guerra
182
182
Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
1
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08
 |Â
show 1 more comment
Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
1
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08
Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
1
1
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
The correct step is as follow
$$cos^2x = 2sin xcos xiff cos^2x - 2sin xcos x=0iff cos x(cos x - 2sin x)=0$$
and therefore the original equation is equivalent to the following $2$ equations
$$cos x=0 quad lor quad cos x - 2sin x=0$$
As an alternative, rephrasing that, we can also observe that
$$cos^2x = 2sin xcos x$$
is clearly satisfied for $cos x=0$ which is a solution then for $cos xneq0$ we can cancel out and obtain
$$cos x = 2sin x$$
Note that the fact is not specifically related to trigonometric function but is a more general fact indeed
$$f(x)cdot g(x)=f(x)cdot h(x)$$
by the same argument is equivalent to the following $2$ equations
$$f(x)=0 quad lor quad g(x)=h(x)$$
answered Sep 7 at 1:06
gimusi
73.9k73889
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
add a comment |Â
up vote
1
down vote
You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:
if $cos x=0$, the equation holds;
else if $cos xne0$, the equation can be reduced to $cos x=2sin x$.
Now you solve the two cases independently.
answered Sep 7 at 9:14
Yves Daoust
115k666209
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The correct step is as follow
$$cos^2x = 2sin xcos xiff cos^2x - 2sin xcos x=0iff cos x(cos x - 2sin x)=0$$
and therefore the original equation is equivalent to the following $2$ equations
$$cos x=0 quad lor quad cos x - 2sin x=0$$
As an alternative, rephrasing that, we can also observe that
$$cos^2x = 2sin xcos x$$
is clearly satisfied for $cos x=0$ which is a solution then for $cos xneq0$ we can cancel out and obtain
$$cos x = 2sin x$$
Note that the fact is not specifically related to trigonometric function but is a more general fact indeed
$$f(x)cdot g(x)=f(x)cdot h(x)$$
by the same argument is equivalent to the following $2$ equations
$$f(x)=0 quad lor quad g(x)=h(x)$$
answered Sep 7 at 1:06
gimusi
73.9k73889
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
add a comment |Â
up vote
7
down vote
accepted
The correct step is as follow
$$cos^2x = 2sin xcos xiff cos^2x - 2sin xcos x=0iff cos x(cos x - 2sin x)=0$$
and therefore the original equation is equivalent to the following $2$ equations
$$cos x=0 quad lor quad cos x - 2sin x=0$$
As an alternative, rephrasing that, we can also observe that
$$cos^2x = 2sin xcos x$$
is clearly satisfied for $cos x=0$ which is a solution then for $cos xneq0$ we can cancel out and obtain
$$cos x = 2sin x$$
Note that the fact is not specifically related to trigonometric function but is a more general fact indeed
$$f(x)cdot g(x)=f(x)cdot h(x)$$
by the same argument is equivalent to the following $2$ equations
$$f(x)=0 quad lor quad g(x)=h(x)$$
answered Sep 7 at 1:06
gimusi
73.9k73889
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The correct step is as follow
$$cos^2x = 2sin xcos xiff cos^2x - 2sin xcos x=0iff cos x(cos x - 2sin x)=0$$
and therefore the original equation is equivalent to the following $2$ equations
$$cos x=0 quad lor quad cos x - 2sin x=0$$
As an alternative, rephrasing that, we can also observe that
$$cos^2x = 2sin xcos x$$
is clearly satisfied for $cos x=0$ which is a solution then for $cos xneq0$ we can cancel out and obtain
$$cos x = 2sin x$$
Note that the fact is not specifically related to trigonometric function but is a more general fact indeed
$$f(x)cdot g(x)=f(x)cdot h(x)$$
by the same argument is equivalent to the following $2$ equations
$$f(x)=0 quad lor quad g(x)=h(x)$$
answered Sep 7 at 1:06
gimusi
73.9k73889
The correct step is as follow
$$cos^2x = 2sin xcos xiff cos^2x - 2sin xcos x=0iff cos x(cos x - 2sin x)=0$$
and therefore the original equation is equivalent to the following $2$ equations
$$cos x=0 quad lor quad cos x - 2sin x=0$$
As an alternative, rephrasing that, we can also observe that
$$cos^2x = 2sin xcos x$$
is clearly satisfied for $cos x=0$ which is a solution then for $cos xneq0$ we can cancel out and obtain
$$cos x = 2sin x$$
Note that the fact is not specifically related to trigonometric function but is a more general fact indeed
$$f(x)cdot g(x)=f(x)cdot h(x)$$
by the same argument is equivalent to the following $2$ equations
$$f(x)=0 quad lor quad g(x)=h(x)$$
answered Sep 7 at 1:06
gimusi
73.9k73889
edited Sep 7 at 9:03
answered Sep 7 at 1:06
gimusi
73.9k73889
answered Sep 7 at 1:06
gimusi
73.9k73889
answered Sep 7 at 1:06
gimusi
73.9k73889
73.9k73889
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
add a comment |Â
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
Yes of course indeed I didn’t write that it is a system but those are 2 independent equations.
– gimusi
Sep 7 at 6:30
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
The equations are not independent: if one is true, then the other is false.
– John Bentin
Sep 7 at 7:50
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
With independent I mean that the final solution is the union of the solutions of the two equations. I fully agree with you.
– gimusi
Sep 7 at 8:00
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Yes, of course we both understand that the solution set is the union of the (disjoint) solution sets of the two equations. But that is not what "two independent equations" means. If you simply write a list of equations, without the very important word "or", the meaning is the conjunction of those equations.
– John Bentin
Sep 7 at 8:18
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
Thanks I fix that point in order to be more clear. Bye
– gimusi
Sep 7 at 8:33
add a comment |Â
up vote
1
down vote
You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:
if $cos x=0$, the equation holds;
else if $cos xne0$, the equation can be reduced to $cos x=2sin x$.
Now you solve the two cases independently.
answered Sep 7 at 9:14
Yves Daoust
115k666209
add a comment |Â
up vote
1
down vote
You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:
if $cos x=0$, the equation holds;
else if $cos xne0$, the equation can be reduced to $cos x=2sin x$.
Now you solve the two cases independently.
answered Sep 7 at 9:14
Yves Daoust
115k666209
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:
if $cos x=0$, the equation holds;
else if $cos xne0$, the equation can be reduced to $cos x=2sin x$.
Now you solve the two cases independently.
answered Sep 7 at 9:14
Yves Daoust
115k666209
You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:
if $cos x=0$, the equation holds;
else if $cos xne0$, the equation can be reduced to $cos x=2sin x$.
Now you solve the two cases independently.
answered Sep 7 at 9:14
Yves Daoust
115k666209
answered Sep 7 at 9:14
Yves Daoust
115k666209
answered Sep 7 at 9:14
Yves Daoust
115k666209
answered Sep 7 at 9:14
Yves Daoust
115k666209
115k666209
add a comment |Â
add a comment |Â
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Dividing out the cos on each side means you are eliminating the solution of pi/2, but you should be able to glean other solutions from what's left. I could be wrong though, I'll try the problem and get back to this!
– Zach
Sep 7 at 0:56
1
There should be no problem dividing by $cos x$, provided you include the solutions to $cos x = 0$.
– lulu
Sep 7 at 0:57
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Sep 7 at 1:01
For the same reason why $,cos xleft( cos x - 2 sin xright)=0,$ is not equivalent to just $,cos x - 2 sin x = 0,$.
– dxiv
Sep 7 at 1:05
You might also find the proof of 1 = 2 to be similar. See math.toronto.edu/mathnet/falseProofs/first1eq2.html
– Biswajit Banerjee
Sep 7 at 1:08