Mixing
How do I solve this equation $ sin(x) = sin(frac{11}{9}pi)$?
$begingroup$
We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it.
$ x = A + k times 2pi$
and
$x = pi - A + k times 2pi$
where $k$ is supposed to be a random integer? and A in this case is $frac{11}{9}pi$
trigonometry
edited Jan 2 at 19:46
Jennifer
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
$endgroup$
add a comment |
$begingroup$
We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it.
$ x = A + k times 2pi$
and
$x = pi - A + k times 2pi$
where $k$ is supposed to be a random integer? and A in this case is $frac{11}{9}pi$
trigonometry
edited Jan 2 at 19:46
Jennifer
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
$endgroup$
1
$begingroup$
Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26
add a comment |
$begingroup$
We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it.
$ x = A + k times 2pi$
and
$x = pi - A + k times 2pi$
where $k$ is supposed to be a random integer? and A in this case is $frac{11}{9}pi$
trigonometry
edited Jan 2 at 19:46
Jennifer
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
$endgroup$
We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it.
$ x = A + k times 2pi$
and
$x = pi - A + k times 2pi$
where $k$ is supposed to be a random integer? and A in this case is $frac{11}{9}pi$
trigonometry
trigonometry
edited Jan 2 at 19:46
Jennifer
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
edited Jan 2 at 19:46
Jennifer
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
edited Jan 2 at 19:46
Jennifer
8,41721737
edited Jan 2 at 19:46
Jennifer
8,41721737
edited Jan 2 at 19:46
Jennifer
8,41721737
8,41721737
asked Jan 2 at 19:20
ythhtrgythhtrg
91
asked Jan 2 at 19:20
ythhtrgythhtrg
91
asked Jan 2 at 19:20
ythhtrgythhtrg
91
91
1
$begingroup$
Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26
add a comment |
1
$begingroup$
Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26
1
1
$begingroup$
Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26
$begingroup$
Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The first step is always the same:
DRAW A DIAGRAM!
It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $sin(x)$ with $x=frac{11}{9}pi$ marked:
$frac{11}{9}pi$ is a bit more than $pi$, so the value of $sin frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.
$x=frac{11}{9}pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2pi$, so $x=frac{11}{9}pi+2pi$ is another solution, and $frac{11}{9}pi+2pi+2pi$ is yet another one, and so forth, and to the other side $frac{11}{9}pi-2pi = -frac{7}{9}pi$ is also a solution, and so forth. This gives us
$$ x = frac{11}{9}pi + 2kpi, kinmathbb Z, $$
the points indicated in green here:
These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2pi$ as our known down crossing is right of $pi$, and we get for this point
$$ x = 2pi - (frac{11}{9}pi - pi) = frac{16}{9}pi $$
This point also has copies at regular $2pi$ intervals to the right, so this gives the blue solutions
$$ x = frac{16}{9}pi + 2kpi, kinmathbb Z $$
The complete solution is $$ x = begin{cases} frac{11}{9}pi + 2kpi & kinmathbb Z \ frac{16}9 pi + 2kpi & k inmathbb Z end{cases} $$
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
$endgroup$
add a comment |
$begingroup$
Note that you have the arcsine function has a range of $left[-frac{pi}{2}, frac{pi}{2}right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $sin(pi-theta) = sin(theta)$.
Hence, we have
$$sin (x) = y implies x = begin{cases} arcsin y+2pi k \ pi-arcsin y+2pi kend{cases}; quad kin mathbb{Z}$$
So, if $$sin(x) = sinleft(frac{11}{9}piright)$$
then there are two cases:
$$x = begin{cases} frac{11}{9}pi+2pi k \ pi-frac{11}{9}pi+2pi kend{cases}$$
with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.
As for why we have $2pi k$ for $k in mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $frac{9pi}{4}$ radians would mean $2pi+frac{pi}{4}$, so it’s represents the same angle as $frac{pi}{4}$. So, any two angles separated by $2pi$, $4pi$, $6pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2pi$ radians or $360°$.
As a simple example, assume we have $sin(x) = 1$. Clearly, the answer must be $pi$. But $pi+2pi$ also works. So does $pi-2pi$. So does any angle $2pi k$ radians apart from $pi$. So we would generalize the answer and give the full solution as $pi+2pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.
However, you may occasionally be given a restricted domain such as $0 leq x leq 2pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2pi k$ for both cases, and in the example I gave, the answer simply becomes $pi$.)
answered Jan 2 at 19:43
KM101KM101
5,9511523
$endgroup$
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
add a comment |
$begingroup$
When you have equation $sin x=a$, the solutions are $x=arcsin a +2pi n$ and $x=pi-arcsin a + 2pi n$. This is based on the identity $sin(pi-a)=sin a$. In your case $a=sin(frac{11pi}{9})$ so the solutions are $x=arcsin(sin(frac{11pi}{9}))+2pi n=-frac{2pi}{9}+2pi n$ and $x=pi-arcsin(sin(frac{11pi}{9}))+2pi n=frac{11pi}{9}+2pi n$. Term $2pi n$ is added becase sine is $2 pi$ periodic thus $sin(a+2pi n)=sin(a)$
answered Jan 2 at 19:36
VasyaVasya
3,2921515
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first step is always the same:
DRAW A DIAGRAM!
It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $sin(x)$ with $x=frac{11}{9}pi$ marked:
$frac{11}{9}pi$ is a bit more than $pi$, so the value of $sin frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.
$x=frac{11}{9}pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2pi$, so $x=frac{11}{9}pi+2pi$ is another solution, and $frac{11}{9}pi+2pi+2pi$ is yet another one, and so forth, and to the other side $frac{11}{9}pi-2pi = -frac{7}{9}pi$ is also a solution, and so forth. This gives us
$$ x = frac{11}{9}pi + 2kpi, kinmathbb Z, $$
the points indicated in green here:
These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2pi$ as our known down crossing is right of $pi$, and we get for this point
$$ x = 2pi - (frac{11}{9}pi - pi) = frac{16}{9}pi $$
This point also has copies at regular $2pi$ intervals to the right, so this gives the blue solutions
$$ x = frac{16}{9}pi + 2kpi, kinmathbb Z $$
The complete solution is $$ x = begin{cases} frac{11}{9}pi + 2kpi & kinmathbb Z \ frac{16}9 pi + 2kpi & k inmathbb Z end{cases} $$
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
$endgroup$
add a comment |
$begingroup$
The first step is always the same:
DRAW A DIAGRAM!
It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $sin(x)$ with $x=frac{11}{9}pi$ marked:
$frac{11}{9}pi$ is a bit more than $pi$, so the value of $sin frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.
$x=frac{11}{9}pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2pi$, so $x=frac{11}{9}pi+2pi$ is another solution, and $frac{11}{9}pi+2pi+2pi$ is yet another one, and so forth, and to the other side $frac{11}{9}pi-2pi = -frac{7}{9}pi$ is also a solution, and so forth. This gives us
$$ x = frac{11}{9}pi + 2kpi, kinmathbb Z, $$
the points indicated in green here:
These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2pi$ as our known down crossing is right of $pi$, and we get for this point
$$ x = 2pi - (frac{11}{9}pi - pi) = frac{16}{9}pi $$
This point also has copies at regular $2pi$ intervals to the right, so this gives the blue solutions
$$ x = frac{16}{9}pi + 2kpi, kinmathbb Z $$
The complete solution is $$ x = begin{cases} frac{11}{9}pi + 2kpi & kinmathbb Z \ frac{16}9 pi + 2kpi & k inmathbb Z end{cases} $$
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
$endgroup$
add a comment |
$begingroup$
The first step is always the same:
DRAW A DIAGRAM!
It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $sin(x)$ with $x=frac{11}{9}pi$ marked:
$frac{11}{9}pi$ is a bit more than $pi$, so the value of $sin frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.
$x=frac{11}{9}pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2pi$, so $x=frac{11}{9}pi+2pi$ is another solution, and $frac{11}{9}pi+2pi+2pi$ is yet another one, and so forth, and to the other side $frac{11}{9}pi-2pi = -frac{7}{9}pi$ is also a solution, and so forth. This gives us
$$ x = frac{11}{9}pi + 2kpi, kinmathbb Z, $$
the points indicated in green here:
These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2pi$ as our known down crossing is right of $pi$, and we get for this point
$$ x = 2pi - (frac{11}{9}pi - pi) = frac{16}{9}pi $$
This point also has copies at regular $2pi$ intervals to the right, so this gives the blue solutions
$$ x = frac{16}{9}pi + 2kpi, kinmathbb Z $$
The complete solution is $$ x = begin{cases} frac{11}{9}pi + 2kpi & kinmathbb Z \ frac{16}9 pi + 2kpi & k inmathbb Z end{cases} $$
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
$endgroup$
The first step is always the same:
DRAW A DIAGRAM!
It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $sin(x)$ with $x=frac{11}{9}pi$ marked:
$frac{11}{9}pi$ is a bit more than $pi$, so the value of $sin frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.
$x=frac{11}{9}pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2pi$, so $x=frac{11}{9}pi+2pi$ is another solution, and $frac{11}{9}pi+2pi+2pi$ is yet another one, and so forth, and to the other side $frac{11}{9}pi-2pi = -frac{7}{9}pi$ is also a solution, and so forth. This gives us
$$ x = frac{11}{9}pi + 2kpi, kinmathbb Z, $$
the points indicated in green here:
These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2pi$ as our known down crossing is right of $pi$, and we get for this point
$$ x = 2pi - (frac{11}{9}pi - pi) = frac{16}{9}pi $$
This point also has copies at regular $2pi$ intervals to the right, so this gives the blue solutions
$$ x = frac{16}{9}pi + 2kpi, kinmathbb Z $$
The complete solution is $$ x = begin{cases} frac{11}{9}pi + 2kpi & kinmathbb Z \ frac{16}9 pi + 2kpi & k inmathbb Z end{cases} $$
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
edited Jan 2 at 20:56
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
answered Jan 2 at 20:40
Henning MakholmHenning Makholm
239k17303540
239k17303540
add a comment |
add a comment |
$begingroup$
Note that you have the arcsine function has a range of $left[-frac{pi}{2}, frac{pi}{2}right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $sin(pi-theta) = sin(theta)$.
Hence, we have
$$sin (x) = y implies x = begin{cases} arcsin y+2pi k \ pi-arcsin y+2pi kend{cases}; quad kin mathbb{Z}$$
So, if $$sin(x) = sinleft(frac{11}{9}piright)$$
then there are two cases:
$$x = begin{cases} frac{11}{9}pi+2pi k \ pi-frac{11}{9}pi+2pi kend{cases}$$
with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.
As for why we have $2pi k$ for $k in mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $frac{9pi}{4}$ radians would mean $2pi+frac{pi}{4}$, so it’s represents the same angle as $frac{pi}{4}$. So, any two angles separated by $2pi$, $4pi$, $6pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2pi$ radians or $360°$.
As a simple example, assume we have $sin(x) = 1$. Clearly, the answer must be $pi$. But $pi+2pi$ also works. So does $pi-2pi$. So does any angle $2pi k$ radians apart from $pi$. So we would generalize the answer and give the full solution as $pi+2pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.
However, you may occasionally be given a restricted domain such as $0 leq x leq 2pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2pi k$ for both cases, and in the example I gave, the answer simply becomes $pi$.)
answered Jan 2 at 19:43
KM101KM101
5,9511523
$endgroup$
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
add a comment |
$begingroup$
Note that you have the arcsine function has a range of $left[-frac{pi}{2}, frac{pi}{2}right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $sin(pi-theta) = sin(theta)$.
Hence, we have
$$sin (x) = y implies x = begin{cases} arcsin y+2pi k \ pi-arcsin y+2pi kend{cases}; quad kin mathbb{Z}$$
So, if $$sin(x) = sinleft(frac{11}{9}piright)$$
then there are two cases:
$$x = begin{cases} frac{11}{9}pi+2pi k \ pi-frac{11}{9}pi+2pi kend{cases}$$
with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.
As for why we have $2pi k$ for $k in mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $frac{9pi}{4}$ radians would mean $2pi+frac{pi}{4}$, so it’s represents the same angle as $frac{pi}{4}$. So, any two angles separated by $2pi$, $4pi$, $6pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2pi$ radians or $360°$.
As a simple example, assume we have $sin(x) = 1$. Clearly, the answer must be $pi$. But $pi+2pi$ also works. So does $pi-2pi$. So does any angle $2pi k$ radians apart from $pi$. So we would generalize the answer and give the full solution as $pi+2pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.
However, you may occasionally be given a restricted domain such as $0 leq x leq 2pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2pi k$ for both cases, and in the example I gave, the answer simply becomes $pi$.)
answered Jan 2 at 19:43
KM101KM101
5,9511523
$endgroup$
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
add a comment |
$begingroup$
Note that you have the arcsine function has a range of $left[-frac{pi}{2}, frac{pi}{2}right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $sin(pi-theta) = sin(theta)$.
Hence, we have
$$sin (x) = y implies x = begin{cases} arcsin y+2pi k \ pi-arcsin y+2pi kend{cases}; quad kin mathbb{Z}$$
So, if $$sin(x) = sinleft(frac{11}{9}piright)$$
then there are two cases:
$$x = begin{cases} frac{11}{9}pi+2pi k \ pi-frac{11}{9}pi+2pi kend{cases}$$
with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.
As for why we have $2pi k$ for $k in mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $frac{9pi}{4}$ radians would mean $2pi+frac{pi}{4}$, so it’s represents the same angle as $frac{pi}{4}$. So, any two angles separated by $2pi$, $4pi$, $6pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2pi$ radians or $360°$.
As a simple example, assume we have $sin(x) = 1$. Clearly, the answer must be $pi$. But $pi+2pi$ also works. So does $pi-2pi$. So does any angle $2pi k$ radians apart from $pi$. So we would generalize the answer and give the full solution as $pi+2pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.
However, you may occasionally be given a restricted domain such as $0 leq x leq 2pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2pi k$ for both cases, and in the example I gave, the answer simply becomes $pi$.)
answered Jan 2 at 19:43
KM101KM101
5,9511523
$endgroup$
Note that you have the arcsine function has a range of $left[-frac{pi}{2}, frac{pi}{2}right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $sin(pi-theta) = sin(theta)$.
Hence, we have
$$sin (x) = y implies x = begin{cases} arcsin y+2pi k \ pi-arcsin y+2pi kend{cases}; quad kin mathbb{Z}$$
So, if $$sin(x) = sinleft(frac{11}{9}piright)$$
then there are two cases:
$$x = begin{cases} frac{11}{9}pi+2pi k \ pi-frac{11}{9}pi+2pi kend{cases}$$
with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.
As for why we have $2pi k$ for $k in mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $frac{9pi}{4}$ radians would mean $2pi+frac{pi}{4}$, so it’s represents the same angle as $frac{pi}{4}$. So, any two angles separated by $2pi$, $4pi$, $6pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2pi$ radians or $360°$.
As a simple example, assume we have $sin(x) = 1$. Clearly, the answer must be $pi$. But $pi+2pi$ also works. So does $pi-2pi$. So does any angle $2pi k$ radians apart from $pi$. So we would generalize the answer and give the full solution as $pi+2pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.
However, you may occasionally be given a restricted domain such as $0 leq x leq 2pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2pi k$ for both cases, and in the example I gave, the answer simply becomes $pi$.)
answered Jan 2 at 19:43
KM101KM101
5,9511523
edited Jan 2 at 20:00
answered Jan 2 at 19:43
KM101KM101
5,9511523
answered Jan 2 at 19:43
KM101KM101
5,9511523
answered Jan 2 at 19:43
KM101KM101
5,9511523
5,9511523
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
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– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
add a comment |
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Thank you, this clarifies a bit but our teacher didn't say anything about arcsin. This is all we got - i.imgur.com/f8BNPwi.png
$endgroup$
– ythhtrg
Jan 2 at 20:11
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Arcsine is the inverse of sine. (I guess you’re given examples which do not involve irregular angles.) You can simply cancel the sines and obtain $frac{11}{9}pi+2pi k$ for the first case and then use the identity $sin(pi-theta) = sin(theta)$ to obtain the second case. I was guessing you’ve studied it, but it’s certainly not required here.
$endgroup$
– KM101
Jan 2 at 20:15
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
$begingroup$
Also, when you have $sin(x) = sin(y)$, the cases simplify to $x = begin{cases} x= y+2pi k \ x = pi-y+2pi k end{cases}$.
$endgroup$
– KM101
Jan 2 at 20:23
add a comment |
$begingroup$
When you have equation $sin x=a$, the solutions are $x=arcsin a +2pi n$ and $x=pi-arcsin a + 2pi n$. This is based on the identity $sin(pi-a)=sin a$. In your case $a=sin(frac{11pi}{9})$ so the solutions are $x=arcsin(sin(frac{11pi}{9}))+2pi n=-frac{2pi}{9}+2pi n$ and $x=pi-arcsin(sin(frac{11pi}{9}))+2pi n=frac{11pi}{9}+2pi n$. Term $2pi n$ is added becase sine is $2 pi$ periodic thus $sin(a+2pi n)=sin(a)$
answered Jan 2 at 19:36
VasyaVasya
3,2921515
$endgroup$
add a comment |
$begingroup$
When you have equation $sin x=a$, the solutions are $x=arcsin a +2pi n$ and $x=pi-arcsin a + 2pi n$. This is based on the identity $sin(pi-a)=sin a$. In your case $a=sin(frac{11pi}{9})$ so the solutions are $x=arcsin(sin(frac{11pi}{9}))+2pi n=-frac{2pi}{9}+2pi n$ and $x=pi-arcsin(sin(frac{11pi}{9}))+2pi n=frac{11pi}{9}+2pi n$. Term $2pi n$ is added becase sine is $2 pi$ periodic thus $sin(a+2pi n)=sin(a)$
answered Jan 2 at 19:36
VasyaVasya
3,2921515
$endgroup$
add a comment |
$begingroup$
When you have equation $sin x=a$, the solutions are $x=arcsin a +2pi n$ and $x=pi-arcsin a + 2pi n$. This is based on the identity $sin(pi-a)=sin a$. In your case $a=sin(frac{11pi}{9})$ so the solutions are $x=arcsin(sin(frac{11pi}{9}))+2pi n=-frac{2pi}{9}+2pi n$ and $x=pi-arcsin(sin(frac{11pi}{9}))+2pi n=frac{11pi}{9}+2pi n$. Term $2pi n$ is added becase sine is $2 pi$ periodic thus $sin(a+2pi n)=sin(a)$
answered Jan 2 at 19:36
VasyaVasya
3,2921515
$endgroup$
When you have equation $sin x=a$, the solutions are $x=arcsin a +2pi n$ and $x=pi-arcsin a + 2pi n$. This is based on the identity $sin(pi-a)=sin a$. In your case $a=sin(frac{11pi}{9})$ so the solutions are $x=arcsin(sin(frac{11pi}{9}))+2pi n=-frac{2pi}{9}+2pi n$ and $x=pi-arcsin(sin(frac{11pi}{9}))+2pi n=frac{11pi}{9}+2pi n$. Term $2pi n$ is added becase sine is $2 pi$ periodic thus $sin(a+2pi n)=sin(a)$
answered Jan 2 at 19:36
VasyaVasya
3,2921515
answered Jan 2 at 19:36
VasyaVasya
3,2921515
answered Jan 2 at 19:36
VasyaVasya
3,2921515
answered Jan 2 at 19:36
VasyaVasya
3,2921515
3,2921515
add a comment |
add a comment |
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Apply mathworld.wolfram.com/ProsthaphaeresisFormulas.html on $sin x=sin11pi/9$
$endgroup$
– lab bhattacharjee
Jan 2 at 19:26