Mixing
What is the value of $operatorname{arctan} left(-frac{12}{5}right)$?
$begingroup$
I am trying to find the value of $sin(operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $operatorname{tan}(theta) = -12/5$. But this is not one of the easy angles to simply look up the value.
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.
trigonometry
edited Jan 28 at 0:55
Zacky
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
$endgroup$
add a comment |
$begingroup$
I am trying to find the value of $sin(operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $operatorname{tan}(theta) = -12/5$. But this is not one of the easy angles to simply look up the value.
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.
trigonometry
edited Jan 28 at 0:55
Zacky
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
$endgroup$
1
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
1
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12
add a comment |
$begingroup$
I am trying to find the value of $sin(operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $operatorname{tan}(theta) = -12/5$. But this is not one of the easy angles to simply look up the value.
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.
trigonometry
edited Jan 28 at 0:55
Zacky
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
$endgroup$
I am trying to find the value of $sin(operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $operatorname{tan}(theta) = -12/5$. But this is not one of the easy angles to simply look up the value.
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.
trigonometry
trigonometry
edited Jan 28 at 0:55
Zacky
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
edited Jan 28 at 0:55
Zacky
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
edited Jan 28 at 0:55
Zacky
7,89511062
edited Jan 28 at 0:55
Zacky
7,89511062
edited Jan 28 at 0:55
Zacky
7,89511062
7,89511062
asked Jan 27 at 22:59
NathanNathan
152
asked Jan 27 at 22:59
NathanNathan
152
asked Jan 27 at 22:59
NathanNathan
152
152
1
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
1
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12
add a comment |
1
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
1
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12
1
1
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
1
1
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$x=operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).
edited Jan 28 at 0:52
Zacky
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
$endgroup$
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
add a comment |
$begingroup$
Let $theta = arctan frac {-12}{5}$
$tan theta = frac {-12}{5} = frac {sin theta}{cos theta}$ which means there there is a right triangle with angle $theta$ and the opposite side = $rsin theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r cos theta = 5$ where $r$ is the hypotenuse.
So $sin theta = frac {-12}r$ where $r$ is the hypotenuse = $sqrt {(-12)^2 + 5^2} = 13$.
So $sin theta = frac {-12}{13}$.
....
In general:
$sin (arcsin frac ab) = frac {a}{sqrt{a^2 + b^2}}$
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
$endgroup$
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$x=operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).
edited Jan 28 at 0:52
Zacky
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
$endgroup$
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
add a comment |
$begingroup$
$x=operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).
edited Jan 28 at 0:52
Zacky
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
$endgroup$
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
add a comment |
$begingroup$
$x=operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).
edited Jan 28 at 0:52
Zacky
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
$endgroup$
$x=operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).
edited Jan 28 at 0:52
Zacky
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
edited Jan 28 at 0:52
Zacky
7,89511062
edited Jan 28 at 0:52
Zacky
7,89511062
edited Jan 28 at 0:52
Zacky
7,89511062
7,89511062
answered Jan 27 at 23:03
GReyesGReyes
2,30815
answered Jan 27 at 23:03
GReyesGReyes
2,30815
answered Jan 27 at 23:03
GReyesGReyes
2,30815
2,30815
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
add a comment |
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
$begingroup$
You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question
$endgroup$
– Nathan
Jan 28 at 1:58
add a comment |
$begingroup$
Let $theta = arctan frac {-12}{5}$
$tan theta = frac {-12}{5} = frac {sin theta}{cos theta}$ which means there there is a right triangle with angle $theta$ and the opposite side = $rsin theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r cos theta = 5$ where $r$ is the hypotenuse.
So $sin theta = frac {-12}r$ where $r$ is the hypotenuse = $sqrt {(-12)^2 + 5^2} = 13$.
So $sin theta = frac {-12}{13}$.
....
In general:
$sin (arcsin frac ab) = frac {a}{sqrt{a^2 + b^2}}$
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
$endgroup$
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
add a comment |
$begingroup$
Let $theta = arctan frac {-12}{5}$
$tan theta = frac {-12}{5} = frac {sin theta}{cos theta}$ which means there there is a right triangle with angle $theta$ and the opposite side = $rsin theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r cos theta = 5$ where $r$ is the hypotenuse.
So $sin theta = frac {-12}r$ where $r$ is the hypotenuse = $sqrt {(-12)^2 + 5^2} = 13$.
So $sin theta = frac {-12}{13}$.
....
In general:
$sin (arcsin frac ab) = frac {a}{sqrt{a^2 + b^2}}$
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
$endgroup$
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
add a comment |
$begingroup$
Let $theta = arctan frac {-12}{5}$
$tan theta = frac {-12}{5} = frac {sin theta}{cos theta}$ which means there there is a right triangle with angle $theta$ and the opposite side = $rsin theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r cos theta = 5$ where $r$ is the hypotenuse.
So $sin theta = frac {-12}r$ where $r$ is the hypotenuse = $sqrt {(-12)^2 + 5^2} = 13$.
So $sin theta = frac {-12}{13}$.
....
In general:
$sin (arcsin frac ab) = frac {a}{sqrt{a^2 + b^2}}$
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
$endgroup$
Let $theta = arctan frac {-12}{5}$
$tan theta = frac {-12}{5} = frac {sin theta}{cos theta}$ which means there there is a right triangle with angle $theta$ and the opposite side = $rsin theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r cos theta = 5$ where $r$ is the hypotenuse.
So $sin theta = frac {-12}r$ where $r$ is the hypotenuse = $sqrt {(-12)^2 + 5^2} = 13$.
So $sin theta = frac {-12}{13}$.
....
In general:
$sin (arcsin frac ab) = frac {a}{sqrt{a^2 + b^2}}$
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
answered Jan 27 at 23:25
fleabloodfleablood
73.4k22891
73.4k22891
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
add a comment |
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
$begingroup$
I suppose that the last expression is $sin (arctan frac ab) = frac {a}{sqrt{a^2 + b^2}}$
$endgroup$
– Claude Leibovici
Jan 28 at 6:08
add a comment |
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1
$begingroup$
If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question.
$endgroup$
– Blue
Jan 27 at 23:04
$begingroup$
This angle is not a rational multiple of $pi$.
$endgroup$
– GEdgar
Jan 28 at 1:35
1
$begingroup$
Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now.
$endgroup$
– Nathan
Jan 28 at 1:56
$begingroup$
@Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need.
$endgroup$
– Blue
Jan 28 at 2:12