How to represent inverse trig function as another inverse trig function?
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I'm struggling with this one. How can I respesent some inverse function as another inverse function? Is it possible to represent let's say $arctan(-2)$ in terms of $arccos(...)$, $arcsin(...)$, $text{arccot(...)}$?
Specific examples:
$arctan{(-2)}$ respresented as $arccos$
$arcsin{(-frac{3}{4}})$ respresented as $arctan$
$arctan{(-3)}$ represented as $arcsin$
I suppose it has to do something with the right triangle, and this is the part that I'm struggling with. I do know how to apply trigonometric functions to the right triangle thing, but no idea how to do it with inverse ones.
Thank you!
real-analysis functions trigonometry
asked Jan 23 at 16:29
wenoweno
29211
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add a comment |
$begingroup$
I'm struggling with this one. How can I respesent some inverse function as another inverse function? Is it possible to represent let's say $arctan(-2)$ in terms of $arccos(...)$, $arcsin(...)$, $text{arccot(...)}$?
Specific examples:
$arctan{(-2)}$ respresented as $arccos$
$arcsin{(-frac{3}{4}})$ respresented as $arctan$
$arctan{(-3)}$ represented as $arcsin$
I suppose it has to do something with the right triangle, and this is the part that I'm struggling with. I do know how to apply trigonometric functions to the right triangle thing, but no idea how to do it with inverse ones.
Thank you!
real-analysis functions trigonometry
asked Jan 23 at 16:29
wenoweno
29211
$endgroup$
$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
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Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
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No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48
add a comment |
$begingroup$
I'm struggling with this one. How can I respesent some inverse function as another inverse function? Is it possible to represent let's say $arctan(-2)$ in terms of $arccos(...)$, $arcsin(...)$, $text{arccot(...)}$?
Specific examples:
$arctan{(-2)}$ respresented as $arccos$
$arcsin{(-frac{3}{4}})$ respresented as $arctan$
$arctan{(-3)}$ represented as $arcsin$
I suppose it has to do something with the right triangle, and this is the part that I'm struggling with. I do know how to apply trigonometric functions to the right triangle thing, but no idea how to do it with inverse ones.
Thank you!
real-analysis functions trigonometry
asked Jan 23 at 16:29
wenoweno
29211
$endgroup$
I'm struggling with this one. How can I respesent some inverse function as another inverse function? Is it possible to represent let's say $arctan(-2)$ in terms of $arccos(...)$, $arcsin(...)$, $text{arccot(...)}$?
Specific examples:
$arctan{(-2)}$ respresented as $arccos$
$arcsin{(-frac{3}{4}})$ respresented as $arctan$
$arctan{(-3)}$ represented as $arcsin$
I suppose it has to do something with the right triangle, and this is the part that I'm struggling with. I do know how to apply trigonometric functions to the right triangle thing, but no idea how to do it with inverse ones.
Thank you!
real-analysis functions trigonometry
real-analysis functions trigonometry
asked Jan 23 at 16:29
wenoweno
29211
asked Jan 23 at 16:29
wenoweno
29211
asked Jan 23 at 16:29
wenoweno
29211
asked Jan 23 at 16:29
wenoweno
29211
asked Jan 23 at 16:29
wenoweno
29211
29211
$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
$begingroup$
Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
$begingroup$
No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48
add a comment |
$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
$begingroup$
Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
$begingroup$
No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48
$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
$begingroup$
Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
$begingroup$
Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
$begingroup$
No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48
$begingroup$
No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48
add a comment |
1 Answer
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You might find these identities useful
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
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$begingroup$
You might find these identities useful
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
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add a comment |
$begingroup$
You might find these identities useful
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
$endgroup$
add a comment |
$begingroup$
You might find these identities useful
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
$endgroup$
You might find these identities useful
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
answered Jan 23 at 17:34
Dr. MathvaDr. Mathva
2,114324
2,114324
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$begingroup$
You might draw a triangle, then everything would be easier.
$endgroup$
– xbh
Jan 23 at 16:31
$begingroup$
Not sure how to apply inverse functions to the triangle.
$endgroup$
– weno
Jan 23 at 16:40
$begingroup$
No, i mean draw a triangle to figure out what angle those inverse trig function represent. E.g. $arctan (-2)$ is an angle whose tangent is $-2$, then it should represent an angle $A $ in a right triangle with the opposed side $-2$ and the adjacent side $1$. Then $cos A$ = adjacent / hypotenuse = $1/sqrt 5$, so it should be $arccos (1/sqrt 5)$ or something. However the image [or, range] of $arccos, arctan$ are different, so you may not be able to do so.
$endgroup$
– xbh
Jan 23 at 16:48