Mixing
Get the linear function without using tangent
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I have to get the linear function of a line on the rectangular coordinate system.
The Line
I know that the line is $ y=tan(90+theta) x+1$ (degree, not RAD)
But is there a way to get the function without using tangent and(or) cotangent?
functions graphing-functions
asked Jul 21 at 3:43
Du Brisingr Arget
305
add a comment |Â
up vote
0
down vote
favorite
I have to get the linear function of a line on the rectangular coordinate system.
The Line
I know that the line is $ y=tan(90+theta) x+1$ (degree, not RAD)
But is there a way to get the function without using tangent and(or) cotangent?
functions graphing-functions
asked Jul 21 at 3:43
Du Brisingr Arget
305
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to get the linear function of a line on the rectangular coordinate system.
The Line
I know that the line is $ y=tan(90+theta) x+1$ (degree, not RAD)
But is there a way to get the function without using tangent and(or) cotangent?
functions graphing-functions
asked Jul 21 at 3:43
Du Brisingr Arget
305
I have to get the linear function of a line on the rectangular coordinate system.
The Line
I know that the line is $ y=tan(90+theta) x+1$ (degree, not RAD)
But is there a way to get the function without using tangent and(or) cotangent?
functions graphing-functions
asked Jul 21 at 3:43
Du Brisingr Arget
305
edited Jul 21 at 5:00
asked Jul 21 at 3:43
Du Brisingr Arget
305
asked Jul 21 at 3:43
Du Brisingr Arget
305
asked Jul 21 at 3:43
Du Brisingr Arget
305
305
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47
add a comment |Â
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
This question is equivalent to asking the question of what the side length of the side of the triangle in the picture on the x-axis is. If you know the equation of the line, you can calculate the y-intercept, which is the length of the side of the triangle on the x-axis, and if you know the side length of the triangle on the x-axis, you can compute the slope and use that formula to find the equation for the line. But the answer to the question "if I have an right triangle with acute angle $theta$ and I want to find the ratio between the opposite and adjacent sides" is nothing more than a definition of what tangent is in the first place, so any solution not in terms of tangent or cotangent is just a derivation of how to express tangent.
More concretely, you might realize, for example, that by definition of cos, the length of the hypotenuse is just $frac1cos(theta)$ and by definition of $sin$, the length of the x-axis side is just $sin(theta)$ times the length of the hypotenuse, so putting this together, you get that the x-axis side of the triangle is $fracsin(theta)cos(theta)$, and hence, the slope of your line is $-fracsin(theta)cos(theta)$, and thus, your line is given by $y = -fracsin(theta)cos(theta) + 1$. But of course, this is just what you discovered in the first place since $fracsincos = tan$.
answered Jul 21 at 5:13
stats_model
55339
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This question is equivalent to asking the question of what the side length of the side of the triangle in the picture on the x-axis is. If you know the equation of the line, you can calculate the y-intercept, which is the length of the side of the triangle on the x-axis, and if you know the side length of the triangle on the x-axis, you can compute the slope and use that formula to find the equation for the line. But the answer to the question "if I have an right triangle with acute angle $theta$ and I want to find the ratio between the opposite and adjacent sides" is nothing more than a definition of what tangent is in the first place, so any solution not in terms of tangent or cotangent is just a derivation of how to express tangent.
More concretely, you might realize, for example, that by definition of cos, the length of the hypotenuse is just $frac1cos(theta)$ and by definition of $sin$, the length of the x-axis side is just $sin(theta)$ times the length of the hypotenuse, so putting this together, you get that the x-axis side of the triangle is $fracsin(theta)cos(theta)$, and hence, the slope of your line is $-fracsin(theta)cos(theta)$, and thus, your line is given by $y = -fracsin(theta)cos(theta) + 1$. But of course, this is just what you discovered in the first place since $fracsincos = tan$.
answered Jul 21 at 5:13
stats_model
55339
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
add a comment |Â
up vote
0
down vote
accepted
This question is equivalent to asking the question of what the side length of the side of the triangle in the picture on the x-axis is. If you know the equation of the line, you can calculate the y-intercept, which is the length of the side of the triangle on the x-axis, and if you know the side length of the triangle on the x-axis, you can compute the slope and use that formula to find the equation for the line. But the answer to the question "if I have an right triangle with acute angle $theta$ and I want to find the ratio between the opposite and adjacent sides" is nothing more than a definition of what tangent is in the first place, so any solution not in terms of tangent or cotangent is just a derivation of how to express tangent.
More concretely, you might realize, for example, that by definition of cos, the length of the hypotenuse is just $frac1cos(theta)$ and by definition of $sin$, the length of the x-axis side is just $sin(theta)$ times the length of the hypotenuse, so putting this together, you get that the x-axis side of the triangle is $fracsin(theta)cos(theta)$, and hence, the slope of your line is $-fracsin(theta)cos(theta)$, and thus, your line is given by $y = -fracsin(theta)cos(theta) + 1$. But of course, this is just what you discovered in the first place since $fracsincos = tan$.
answered Jul 21 at 5:13
stats_model
55339
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This question is equivalent to asking the question of what the side length of the side of the triangle in the picture on the x-axis is. If you know the equation of the line, you can calculate the y-intercept, which is the length of the side of the triangle on the x-axis, and if you know the side length of the triangle on the x-axis, you can compute the slope and use that formula to find the equation for the line. But the answer to the question "if I have an right triangle with acute angle $theta$ and I want to find the ratio between the opposite and adjacent sides" is nothing more than a definition of what tangent is in the first place, so any solution not in terms of tangent or cotangent is just a derivation of how to express tangent.
More concretely, you might realize, for example, that by definition of cos, the length of the hypotenuse is just $frac1cos(theta)$ and by definition of $sin$, the length of the x-axis side is just $sin(theta)$ times the length of the hypotenuse, so putting this together, you get that the x-axis side of the triangle is $fracsin(theta)cos(theta)$, and hence, the slope of your line is $-fracsin(theta)cos(theta)$, and thus, your line is given by $y = -fracsin(theta)cos(theta) + 1$. But of course, this is just what you discovered in the first place since $fracsincos = tan$.
answered Jul 21 at 5:13
stats_model
55339
This question is equivalent to asking the question of what the side length of the side of the triangle in the picture on the x-axis is. If you know the equation of the line, you can calculate the y-intercept, which is the length of the side of the triangle on the x-axis, and if you know the side length of the triangle on the x-axis, you can compute the slope and use that formula to find the equation for the line. But the answer to the question "if I have an right triangle with acute angle $theta$ and I want to find the ratio between the opposite and adjacent sides" is nothing more than a definition of what tangent is in the first place, so any solution not in terms of tangent or cotangent is just a derivation of how to express tangent.
More concretely, you might realize, for example, that by definition of cos, the length of the hypotenuse is just $frac1cos(theta)$ and by definition of $sin$, the length of the x-axis side is just $sin(theta)$ times the length of the hypotenuse, so putting this together, you get that the x-axis side of the triangle is $fracsin(theta)cos(theta)$, and hence, the slope of your line is $-fracsin(theta)cos(theta)$, and thus, your line is given by $y = -fracsin(theta)cos(theta) + 1$. But of course, this is just what you discovered in the first place since $fracsincos = tan$.
answered Jul 21 at 5:13
stats_model
55339
edited Jul 25 at 22:31
answered Jul 21 at 5:13
stats_model
55339
answered Jul 21 at 5:13
stats_model
55339
answered Jul 21 at 5:13
stats_model
55339
55339
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
add a comment |Â
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
Thanks I think you're right, but I'll wait a few more days to see if someone(though not likely) has a different answer
– Du Brisingr Arget
Jul 21 at 15:55
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858231%2fget-the-linear-function-without-using-tangent%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Unless you know $theta$ has a special value (like $pi/4 = 45^circ$ for example) then you cannot simplify it further.
– Winther
Jul 21 at 3:47