How to find the intersect point between two sine waves?
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I'm trying to find the points where (y = 2 sin x) and (x = 2 sin y) intersect, besides (0, 0).
I've managed to work out that 2 sin x should equal x.
My classpad can give me an answer that I believe is right, simply by looking at these two sine functions on a graph. But how can I go about solving this problem?
Thanks in advance.
trigonometry
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add a comment |
$begingroup$
I'm trying to find the points where (y = 2 sin x) and (x = 2 sin y) intersect, besides (0, 0).
I've managed to work out that 2 sin x should equal x.
My classpad can give me an answer that I believe is right, simply by looking at these two sine functions on a graph. But how can I go about solving this problem?
Thanks in advance.
trigonometry
$endgroup$
add a comment |
$begingroup$
I'm trying to find the points where (y = 2 sin x) and (x = 2 sin y) intersect, besides (0, 0).
I've managed to work out that 2 sin x should equal x.
My classpad can give me an answer that I believe is right, simply by looking at these two sine functions on a graph. But how can I go about solving this problem?
Thanks in advance.
trigonometry
$endgroup$
I'm trying to find the points where (y = 2 sin x) and (x = 2 sin y) intersect, besides (0, 0).
I've managed to work out that 2 sin x should equal x.
My classpad can give me an answer that I believe is right, simply by looking at these two sine functions on a graph. But how can I go about solving this problem?
Thanks in advance.
trigonometry
trigonometry
asked May 6 '17 at 10:10
Three OneFourThree OneFour
1099
1099
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add a comment |
2 Answers
2
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$begingroup$
plugging $$x=2sin(y)$$ into your first equation you will get
$$y=2sin(2sin(y))$$ this equation can only be solved by a numerical way,e.g. the Newton-Raphson method. one solution is given by $$1.895494267$$
$endgroup$
add a comment |
$begingroup$
You can find the numerical value of the intersection with a common scientific calculator repeatedly calculating "sin" (take care that trigonometric functions are set to "rad") and multiplying the result by $2$ until the result stabilizes (at each iteration you get the same value). You can choose any starting point between $0$ and $pi$ (but staying close to $1.9$ will make the process shorter). You are in fact looking for the attractor point of the iterated function $y=2sin x$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
plugging $$x=2sin(y)$$ into your first equation you will get
$$y=2sin(2sin(y))$$ this equation can only be solved by a numerical way,e.g. the Newton-Raphson method. one solution is given by $$1.895494267$$
$endgroup$
add a comment |
$begingroup$
plugging $$x=2sin(y)$$ into your first equation you will get
$$y=2sin(2sin(y))$$ this equation can only be solved by a numerical way,e.g. the Newton-Raphson method. one solution is given by $$1.895494267$$
$endgroup$
add a comment |
$begingroup$
plugging $$x=2sin(y)$$ into your first equation you will get
$$y=2sin(2sin(y))$$ this equation can only be solved by a numerical way,e.g. the Newton-Raphson method. one solution is given by $$1.895494267$$
$endgroup$
plugging $$x=2sin(y)$$ into your first equation you will get
$$y=2sin(2sin(y))$$ this equation can only be solved by a numerical way,e.g. the Newton-Raphson method. one solution is given by $$1.895494267$$
answered May 6 '17 at 10:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.4k42865
74.4k42865
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$begingroup$
You can find the numerical value of the intersection with a common scientific calculator repeatedly calculating "sin" (take care that trigonometric functions are set to "rad") and multiplying the result by $2$ until the result stabilizes (at each iteration you get the same value). You can choose any starting point between $0$ and $pi$ (but staying close to $1.9$ will make the process shorter). You are in fact looking for the attractor point of the iterated function $y=2sin x$.
$endgroup$
add a comment |
$begingroup$
You can find the numerical value of the intersection with a common scientific calculator repeatedly calculating "sin" (take care that trigonometric functions are set to "rad") and multiplying the result by $2$ until the result stabilizes (at each iteration you get the same value). You can choose any starting point between $0$ and $pi$ (but staying close to $1.9$ will make the process shorter). You are in fact looking for the attractor point of the iterated function $y=2sin x$.
$endgroup$
add a comment |
$begingroup$
You can find the numerical value of the intersection with a common scientific calculator repeatedly calculating "sin" (take care that trigonometric functions are set to "rad") and multiplying the result by $2$ until the result stabilizes (at each iteration you get the same value). You can choose any starting point between $0$ and $pi$ (but staying close to $1.9$ will make the process shorter). You are in fact looking for the attractor point of the iterated function $y=2sin x$.
$endgroup$
You can find the numerical value of the intersection with a common scientific calculator repeatedly calculating "sin" (take care that trigonometric functions are set to "rad") and multiplying the result by $2$ until the result stabilizes (at each iteration you get the same value). You can choose any starting point between $0$ and $pi$ (but staying close to $1.9$ will make the process shorter). You are in fact looking for the attractor point of the iterated function $y=2sin x$.
answered May 6 '17 at 13:02
lesath82lesath82
1,921226
1,921226
add a comment |
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