Mixing
Confusion with Finding Solutions to Trigonometric Problems within a Range
$begingroup$
I have been having problems recently concerning the solutions that are presented to trigonometric problems. For example, there is a case such as 2 + 3sin 2x = 0. With all trigonometric problems that are in some form or another, I know that I will have to apply algebra or trigonometric identities to isolate x or theta or to isolate a single trigonometric function with a single angle equating to some value on the right hand side before finding all the solutions. However, I am puzzled by the answers provided for many trigonometric problems. In the example stated above, it is restricted to the range of x is less than or equal to 180 and greater than or equal to 0. Solving for x, I obtain -20.9 degrees. I see that this is out of the range, but if I take the mirroring amplitude of this answer (+0.36), I see that a line drawn on the graph of the sin function intersects the function at two points, one is at +20.9 degrees and the other at 159.1 (180 - 20.9) degrees. But it is stated in my textbook that 110.9 degrees is also the answer.
Because of this, I would appreciate any help that would assist me in understanding how to find all the solutions to a problem within a restricted function. I have been solving many of these trigonometric problems, but when I get to the part where I have to state the solutions that satisfy such problems I am completely lost. I have been watching several videos and reading online tutorials about this subject matter in mathematics, but I am not able to see their reasoning for multiple solutions.
trigonometry
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
$endgroup$
add a comment |
$begingroup$
I have been having problems recently concerning the solutions that are presented to trigonometric problems. For example, there is a case such as 2 + 3sin 2x = 0. With all trigonometric problems that are in some form or another, I know that I will have to apply algebra or trigonometric identities to isolate x or theta or to isolate a single trigonometric function with a single angle equating to some value on the right hand side before finding all the solutions. However, I am puzzled by the answers provided for many trigonometric problems. In the example stated above, it is restricted to the range of x is less than or equal to 180 and greater than or equal to 0. Solving for x, I obtain -20.9 degrees. I see that this is out of the range, but if I take the mirroring amplitude of this answer (+0.36), I see that a line drawn on the graph of the sin function intersects the function at two points, one is at +20.9 degrees and the other at 159.1 (180 - 20.9) degrees. But it is stated in my textbook that 110.9 degrees is also the answer.
Because of this, I would appreciate any help that would assist me in understanding how to find all the solutions to a problem within a restricted function. I have been solving many of these trigonometric problems, but when I get to the part where I have to state the solutions that satisfy such problems I am completely lost. I have been watching several videos and reading online tutorials about this subject matter in mathematics, but I am not able to see their reasoning for multiple solutions.
trigonometry
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
$endgroup$
$begingroup$
I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22
add a comment |
$begingroup$
I have been having problems recently concerning the solutions that are presented to trigonometric problems. For example, there is a case such as 2 + 3sin 2x = 0. With all trigonometric problems that are in some form or another, I know that I will have to apply algebra or trigonometric identities to isolate x or theta or to isolate a single trigonometric function with a single angle equating to some value on the right hand side before finding all the solutions. However, I am puzzled by the answers provided for many trigonometric problems. In the example stated above, it is restricted to the range of x is less than or equal to 180 and greater than or equal to 0. Solving for x, I obtain -20.9 degrees. I see that this is out of the range, but if I take the mirroring amplitude of this answer (+0.36), I see that a line drawn on the graph of the sin function intersects the function at two points, one is at +20.9 degrees and the other at 159.1 (180 - 20.9) degrees. But it is stated in my textbook that 110.9 degrees is also the answer.
Because of this, I would appreciate any help that would assist me in understanding how to find all the solutions to a problem within a restricted function. I have been solving many of these trigonometric problems, but when I get to the part where I have to state the solutions that satisfy such problems I am completely lost. I have been watching several videos and reading online tutorials about this subject matter in mathematics, but I am not able to see their reasoning for multiple solutions.
trigonometry
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
$endgroup$
I have been having problems recently concerning the solutions that are presented to trigonometric problems. For example, there is a case such as 2 + 3sin 2x = 0. With all trigonometric problems that are in some form or another, I know that I will have to apply algebra or trigonometric identities to isolate x or theta or to isolate a single trigonometric function with a single angle equating to some value on the right hand side before finding all the solutions. However, I am puzzled by the answers provided for many trigonometric problems. In the example stated above, it is restricted to the range of x is less than or equal to 180 and greater than or equal to 0. Solving for x, I obtain -20.9 degrees. I see that this is out of the range, but if I take the mirroring amplitude of this answer (+0.36), I see that a line drawn on the graph of the sin function intersects the function at two points, one is at +20.9 degrees and the other at 159.1 (180 - 20.9) degrees. But it is stated in my textbook that 110.9 degrees is also the answer.
Because of this, I would appreciate any help that would assist me in understanding how to find all the solutions to a problem within a restricted function. I have been solving many of these trigonometric problems, but when I get to the part where I have to state the solutions that satisfy such problems I am completely lost. I have been watching several videos and reading online tutorials about this subject matter in mathematics, but I am not able to see their reasoning for multiple solutions.
trigonometry
trigonometry
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
edited Jan 1 at 16:21
The Contextual Path
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
asked Jan 1 at 13:34
The Contextual PathThe Contextual Path
155
155
$begingroup$
I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22
add a comment |
$begingroup$
I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22
$begingroup$
I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $cos 2x = -frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-frac 23$ (a negative number)
2) it lies in the correct range. Since $0 leq x leq 180^{circ}, 0 leq y leq 360^{circ}$
The range $0 leq y leq 360^{circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{circ} leq y leq 270^{circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $frac 23$. Note that you can ignore the sign here. You'll get $arccos frac 23 approx 48.19^{circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{circ}$, i.e find $180^{circ}- 48.19^{circ} = 131.81^{circ}$. To get an answer in the third quadrant, add $180^{circ}$ to it, i.e. find $180^{circ}+ 48.19^{circ} = 228.19^{circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{circ}$ and $114.09^{circ}$, which you can verify by substituting back into the original equation.
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
$endgroup$
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
add a comment |
$begingroup$
It's important to keep in mind that trigonometric functions are periodic.
$$sin(x) = sin(x + 2kpi)$$
$$cos(x) = sin(x + 2kpi)$$
$$tan(x) = tan(x + k pi)$$
All of the above hold for $k$ being an integer. (If you're preferring degrees, replace $pi$ with $180^circ$.)
Another important property is that sine/tangent are odd functions, and cosine is even:
$$sin(-x) = -sin(x)$$
$$cos(-x) = cos(x)$$
$$tan(-x) = -tan(x)$$
(This is, of course, not touching on that your answers were wrong in the first place, which others have already pointed other.)
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
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2 Answers
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$begingroup$
Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $cos 2x = -frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-frac 23$ (a negative number)
2) it lies in the correct range. Since $0 leq x leq 180^{circ}, 0 leq y leq 360^{circ}$
The range $0 leq y leq 360^{circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{circ} leq y leq 270^{circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $frac 23$. Note that you can ignore the sign here. You'll get $arccos frac 23 approx 48.19^{circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{circ}$, i.e find $180^{circ}- 48.19^{circ} = 131.81^{circ}$. To get an answer in the third quadrant, add $180^{circ}$ to it, i.e. find $180^{circ}+ 48.19^{circ} = 228.19^{circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{circ}$ and $114.09^{circ}$, which you can verify by substituting back into the original equation.
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
$endgroup$
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
add a comment |
$begingroup$
Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $cos 2x = -frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-frac 23$ (a negative number)
2) it lies in the correct range. Since $0 leq x leq 180^{circ}, 0 leq y leq 360^{circ}$
The range $0 leq y leq 360^{circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{circ} leq y leq 270^{circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $frac 23$. Note that you can ignore the sign here. You'll get $arccos frac 23 approx 48.19^{circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{circ}$, i.e find $180^{circ}- 48.19^{circ} = 131.81^{circ}$. To get an answer in the third quadrant, add $180^{circ}$ to it, i.e. find $180^{circ}+ 48.19^{circ} = 228.19^{circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{circ}$ and $114.09^{circ}$, which you can verify by substituting back into the original equation.
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
$endgroup$
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
add a comment |
$begingroup$
Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $cos 2x = -frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-frac 23$ (a negative number)
2) it lies in the correct range. Since $0 leq x leq 180^{circ}, 0 leq y leq 360^{circ}$
The range $0 leq y leq 360^{circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{circ} leq y leq 270^{circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $frac 23$. Note that you can ignore the sign here. You'll get $arccos frac 23 approx 48.19^{circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{circ}$, i.e find $180^{circ}- 48.19^{circ} = 131.81^{circ}$. To get an answer in the third quadrant, add $180^{circ}$ to it, i.e. find $180^{circ}+ 48.19^{circ} = 228.19^{circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{circ}$ and $114.09^{circ}$, which you can verify by substituting back into the original equation.
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
$endgroup$
Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $cos 2x = -frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-frac 23$ (a negative number)
2) it lies in the correct range. Since $0 leq x leq 180^{circ}, 0 leq y leq 360^{circ}$
The range $0 leq y leq 360^{circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{circ} leq y leq 270^{circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $frac 23$. Note that you can ignore the sign here. You'll get $arccos frac 23 approx 48.19^{circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{circ}$, i.e find $180^{circ}- 48.19^{circ} = 131.81^{circ}$. To get an answer in the third quadrant, add $180^{circ}$ to it, i.e. find $180^{circ}+ 48.19^{circ} = 228.19^{circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{circ}$ and $114.09^{circ}$, which you can verify by substituting back into the original equation.
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
edited Jan 1 at 14:32
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
answered Jan 1 at 14:08
DeepakDeepak
16.8k11436
16.8k11436
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
add a comment |
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
1
1
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
$begingroup$
Although I have typed in the equation wrongly, as I meant to say sine instead of cosine, your post has provided me with some clarity as to how to go about solving trigonometric problems. Thank you very much for that. I will try to solve some problems now and I will get back to you.
$endgroup$
– The Contextual Path
Jan 1 at 16:24
add a comment |
$begingroup$
It's important to keep in mind that trigonometric functions are periodic.
$$sin(x) = sin(x + 2kpi)$$
$$cos(x) = sin(x + 2kpi)$$
$$tan(x) = tan(x + k pi)$$
All of the above hold for $k$ being an integer. (If you're preferring degrees, replace $pi$ with $180^circ$.)
Another important property is that sine/tangent are odd functions, and cosine is even:
$$sin(-x) = -sin(x)$$
$$cos(-x) = cos(x)$$
$$tan(-x) = -tan(x)$$
(This is, of course, not touching on that your answers were wrong in the first place, which others have already pointed other.)
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
$endgroup$
add a comment |
$begingroup$
It's important to keep in mind that trigonometric functions are periodic.
$$sin(x) = sin(x + 2kpi)$$
$$cos(x) = sin(x + 2kpi)$$
$$tan(x) = tan(x + k pi)$$
All of the above hold for $k$ being an integer. (If you're preferring degrees, replace $pi$ with $180^circ$.)
Another important property is that sine/tangent are odd functions, and cosine is even:
$$sin(-x) = -sin(x)$$
$$cos(-x) = cos(x)$$
$$tan(-x) = -tan(x)$$
(This is, of course, not touching on that your answers were wrong in the first place, which others have already pointed other.)
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
$endgroup$
add a comment |
$begingroup$
It's important to keep in mind that trigonometric functions are periodic.
$$sin(x) = sin(x + 2kpi)$$
$$cos(x) = sin(x + 2kpi)$$
$$tan(x) = tan(x + k pi)$$
All of the above hold for $k$ being an integer. (If you're preferring degrees, replace $pi$ with $180^circ$.)
Another important property is that sine/tangent are odd functions, and cosine is even:
$$sin(-x) = -sin(x)$$
$$cos(-x) = cos(x)$$
$$tan(-x) = -tan(x)$$
(This is, of course, not touching on that your answers were wrong in the first place, which others have already pointed other.)
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
$endgroup$
It's important to keep in mind that trigonometric functions are periodic.
$$sin(x) = sin(x + 2kpi)$$
$$cos(x) = sin(x + 2kpi)$$
$$tan(x) = tan(x + k pi)$$
All of the above hold for $k$ being an integer. (If you're preferring degrees, replace $pi$ with $180^circ$.)
Another important property is that sine/tangent are odd functions, and cosine is even:
$$sin(-x) = -sin(x)$$
$$cos(-x) = cos(x)$$
$$tan(-x) = -tan(x)$$
(This is, of course, not touching on that your answers were wrong in the first place, which others have already pointed other.)
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
edited Jan 1 at 17:35
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
answered Jan 1 at 13:46
Eevee TrainerEevee Trainer
5,3531836
5,3531836
add a comment |
add a comment |
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I am very confused as to how you got $20.9^circ$ as a solution to $2+3cos(2x)=0$. Personally, I am getting the solutions $65.9^circ$ and $114.1^circ$. It would really help us if you could show us your work so we could see where you went wrong.
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– Noble Mushtak
Jan 1 at 13:42
$begingroup$
I am very sorry, I had typed in cosine when I meant to say sine.
$endgroup$
– The Contextual Path
Jan 1 at 16:22