Mathematical solution for intuitive trigonometric equation
$begingroup$
I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:
$sin 3alpha = 2sinalpha$
Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:
$sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$
Therefore, $alpha = 30^circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.
How can I solve this type of problem when the intuitive approach fails?
trigonometry
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:
$sin 3alpha = 2sinalpha$
Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:
$sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$
Therefore, $alpha = 30^circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.
How can I solve this type of problem when the intuitive approach fails?
trigonometry
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
$endgroup$
add a comment |
$begingroup$
I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:
$sin 3alpha = 2sinalpha$
Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:
$sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$
Therefore, $alpha = 30^circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.
How can I solve this type of problem when the intuitive approach fails?
trigonometry
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
$endgroup$
I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:
$sin 3alpha = 2sinalpha$
Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:
$sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$
Therefore, $alpha = 30^circ$ is a possible solution.
I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.
How can I solve this type of problem when the intuitive approach fails?
trigonometry
trigonometry
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
edited Dec 8 '18 at 15:32
daedsidog
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
asked Dec 8 '18 at 15:26
daedsidogdaedsidog
29017
29017
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If triple angle relation for sine is known, letting $ sin alpha =s$,
$$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
$endgroup$
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
add a comment |
$begingroup$
Here is a lead.
$$sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x).$$
Now invoke some double angle magig to get
$$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
Next, use the pythagorean identities to gete
$$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
Can you use this?
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
$endgroup$
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If triple angle relation for sine is known, letting $ sin alpha =s$,
$$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
$endgroup$
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
add a comment |
$begingroup$
If triple angle relation for sine is known, letting $ sin alpha =s$,
$$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
$endgroup$
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
add a comment |
$begingroup$
If triple angle relation for sine is known, letting $ sin alpha =s$,
$$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
$endgroup$
If triple angle relation for sine is known, letting $ sin alpha =s$,
$$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
edited Dec 8 '18 at 16:14
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
answered Dec 8 '18 at 15:37
NarasimhamNarasimham
20.7k52158
20.7k52158
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
add a comment |
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Forgive my ignorance, but I don't understand what this syntax is meant to represent.
$endgroup$
– daedsidog
Dec 8 '18 at 15:48
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
$begingroup$
Sorry I should have given it fully, shall edit it in the answer.
$endgroup$
– Narasimham
Dec 8 '18 at 16:14
add a comment |
$begingroup$
Here is a lead.
$$sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x).$$
Now invoke some double angle magig to get
$$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
Next, use the pythagorean identities to gete
$$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
Can you use this?
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
$endgroup$
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
add a comment |
$begingroup$
Here is a lead.
$$sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x).$$
Now invoke some double angle magig to get
$$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
Next, use the pythagorean identities to gete
$$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
Can you use this?
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
$endgroup$
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
add a comment |
$begingroup$
Here is a lead.
$$sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x).$$
Now invoke some double angle magig to get
$$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
Next, use the pythagorean identities to gete
$$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
Can you use this?
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
$endgroup$
Here is a lead.
$$sin(3x) = sin(2x + x)
= sin(2x)cos(x) + cos(2x)sin(x).$$
Now invoke some double angle magig to get
$$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
Next, use the pythagorean identities to gete
$$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
Can you use this?
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
answered Dec 8 '18 at 15:39
ncmathsadistncmathsadist
42.7k260103
42.7k260103
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
add a comment |
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
$begingroup$
This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
$endgroup$
– daedsidog
Dec 8 '18 at 15:47
add a comment |
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