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Find the values of $x$ and $y$ that satisfies $sin(x+y)=sin x+sin y$.
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I know that in general the following equality does not hold: $sin(x+y)=sin x + sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$sin(x+y)=sin xcos y + sin y cos x = sin x + sin y$. Then from this equation, I got
$sin x(1-cos y)+sin y(1-cos x)=0$.
There are two possibilities:
Case 1: $sin x(1-cos y)=0$ and $sin y(1-cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2pi k$ for some integer $k$.
Case 2: $sin x(1-cos y)=n$ and $sin y(1-cos x)=-n$ where $n$ is a nonzero real number. Then we have
$sin x(1-cos y)=sin y(cos x-1)$ Since $nneq0$ then we can divide both sides of this equation by $sin x sin y)$ to obtain $csc y - cot y = cot x - csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
trigonometry
asked Aug 3 at 18:30
user573497
1908
add a comment |Â
up vote
4
down vote
favorite
I know that in general the following equality does not hold: $sin(x+y)=sin x + sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$sin(x+y)=sin xcos y + sin y cos x = sin x + sin y$. Then from this equation, I got
$sin x(1-cos y)+sin y(1-cos x)=0$.
There are two possibilities:
Case 1: $sin x(1-cos y)=0$ and $sin y(1-cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2pi k$ for some integer $k$.
Case 2: $sin x(1-cos y)=n$ and $sin y(1-cos x)=-n$ where $n$ is a nonzero real number. Then we have
$sin x(1-cos y)=sin y(cos x-1)$ Since $nneq0$ then we can divide both sides of this equation by $sin x sin y)$ to obtain $csc y - cot y = cot x - csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
trigonometry
asked Aug 3 at 18:30
user573497
1908
Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I know that in general the following equality does not hold: $sin(x+y)=sin x + sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$sin(x+y)=sin xcos y + sin y cos x = sin x + sin y$. Then from this equation, I got
$sin x(1-cos y)+sin y(1-cos x)=0$.
There are two possibilities:
Case 1: $sin x(1-cos y)=0$ and $sin y(1-cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2pi k$ for some integer $k$.
Case 2: $sin x(1-cos y)=n$ and $sin y(1-cos x)=-n$ where $n$ is a nonzero real number. Then we have
$sin x(1-cos y)=sin y(cos x-1)$ Since $nneq0$ then we can divide both sides of this equation by $sin x sin y)$ to obtain $csc y - cot y = cot x - csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
trigonometry
asked Aug 3 at 18:30
user573497
1908
I know that in general the following equality does not hold: $sin(x+y)=sin x + sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$sin(x+y)=sin xcos y + sin y cos x = sin x + sin y$. Then from this equation, I got
$sin x(1-cos y)+sin y(1-cos x)=0$.
There are two possibilities:
Case 1: $sin x(1-cos y)=0$ and $sin y(1-cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2pi k$ for some integer $k$.
Case 2: $sin x(1-cos y)=n$ and $sin y(1-cos x)=-n$ where $n$ is a nonzero real number. Then we have
$sin x(1-cos y)=sin y(cos x-1)$ Since $nneq0$ then we can divide both sides of this equation by $sin x sin y)$ to obtain $csc y - cot y = cot x - csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
trigonometry
asked Aug 3 at 18:30
user573497
1908
asked Aug 3 at 18:30
user573497
1908
asked Aug 3 at 18:30
user573497
1908
asked Aug 3 at 18:30
user573497
1908
1908
Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35
add a comment |Â
Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35
Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
We know that
$$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
and by sum to product formula we have
$$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
therefore
$$sin(x+y)=sin(x)+sin(y) $$
$$iff 2sinleft(fracx+y2right)cosleft(fracx+y2right)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
then we need to consider two cases
$$2sinleft(fracx+y2right)=0 $$
or otherwise we can cancel out $2sinleft(fracx+y2right)$ and obtain
$$cosleft(fracx+y2right)=cosleft(fracx-y2right)$$
answered Aug 3 at 18:48
gimusi
63.7k73480
add a comment |Â
up vote
8
down vote
use that $$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$ and $$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
that's the key idea!
– gimusi
Aug 3 at 18:57
add a comment |Â
up vote
0
down vote
Try $$x=ypm2kpi$$ and $$x=-ypm 2kpi$$
See if these are the only solutions.
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
We know that
$$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
and by sum to product formula we have
$$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
therefore
$$sin(x+y)=sin(x)+sin(y) $$
$$iff 2sinleft(fracx+y2right)cosleft(fracx+y2right)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
then we need to consider two cases
$$2sinleft(fracx+y2right)=0 $$
or otherwise we can cancel out $2sinleft(fracx+y2right)$ and obtain
$$cosleft(fracx+y2right)=cosleft(fracx-y2right)$$
answered Aug 3 at 18:48
gimusi
63.7k73480
add a comment |Â
up vote
4
down vote
accepted
We know that
$$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
and by sum to product formula we have
$$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
therefore
$$sin(x+y)=sin(x)+sin(y) $$
$$iff 2sinleft(fracx+y2right)cosleft(fracx+y2right)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
then we need to consider two cases
$$2sinleft(fracx+y2right)=0 $$
or otherwise we can cancel out $2sinleft(fracx+y2right)$ and obtain
$$cosleft(fracx+y2right)=cosleft(fracx-y2right)$$
answered Aug 3 at 18:48
gimusi
63.7k73480
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
We know that
$$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
and by sum to product formula we have
$$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
therefore
$$sin(x+y)=sin(x)+sin(y) $$
$$iff 2sinleft(fracx+y2right)cosleft(fracx+y2right)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
then we need to consider two cases
$$2sinleft(fracx+y2right)=0 $$
or otherwise we can cancel out $2sinleft(fracx+y2right)$ and obtain
$$cosleft(fracx+y2right)=cosleft(fracx-y2right)$$
answered Aug 3 at 18:48
gimusi
63.7k73480
We know that
$$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
and by sum to product formula we have
$$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
therefore
$$sin(x+y)=sin(x)+sin(y) $$
$$iff 2sinleft(fracx+y2right)cosleft(fracx+y2right)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$
then we need to consider two cases
$$2sinleft(fracx+y2right)=0 $$
or otherwise we can cancel out $2sinleft(fracx+y2right)$ and obtain
$$cosleft(fracx+y2right)=cosleft(fracx-y2right)$$
answered Aug 3 at 18:48
gimusi
63.7k73480
edited Aug 3 at 18:53
answered Aug 3 at 18:48
gimusi
63.7k73480
answered Aug 3 at 18:48
gimusi
63.7k73480
answered Aug 3 at 18:48
gimusi
63.7k73480
63.7k73480
add a comment |Â
add a comment |Â
up vote
8
down vote
use that $$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$ and $$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
that's the key idea!
– gimusi
Aug 3 at 18:57
add a comment |Â
up vote
8
down vote
use that $$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$ and $$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
that's the key idea!
– gimusi
Aug 3 at 18:57
add a comment |Â
up vote
8
down vote
up vote
8
down vote
use that $$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$ and $$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
use that $$sin(x)+sin(y)=2sinleft(fracx+y2right)cosleft(fracx-y2right)$$ and $$sin(x+y)=2sinleft(fracx+y2right)cosleft(fracx+y2right)$$
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:42
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
that's the key idea!
– gimusi
Aug 3 at 18:57
add a comment |Â
that's the key idea!
– gimusi
Aug 3 at 18:57
that's the key idea!
– gimusi
Aug 3 at 18:57
that's the key idea!
– gimusi
Aug 3 at 18:57
add a comment |Â
up vote
0
down vote
Try $$x=ypm2kpi$$ and $$x=-ypm 2kpi$$
See if these are the only solutions.
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
0
down vote
Try $$x=ypm2kpi$$ and $$x=-ypm 2kpi$$
See if these are the only solutions.
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Try $$x=ypm2kpi$$ and $$x=-ypm 2kpi$$
See if these are the only solutions.
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
Try $$x=ypm2kpi$$ and $$x=-ypm 2kpi$$
See if these are the only solutions.
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
answered Aug 3 at 18:50
Mohammad Riazi-Kermani
27k41850
27k41850
add a comment |Â
add a comment |Â
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Take a look at the convexity of the sine function, that may be helpful.
– Laz
Aug 3 at 18:34
In second case use $1-cos 2A=2sin^2A$ and $sin 2A=2sin A cos A$. That can help.
– Anurag A
Aug 3 at 18:35