Mixing
Evaluating $sec^{-1}left(frac{-2}{sqrt{3}}right)$ and $sec^{-1}left(-sqrt{2}right)$. Why are my answers...
$begingroup$
$$sec^{-1}left(frac{-2}{sqrt{3}}right)$$
$$sec^{-1}left(-sqrt{2}right)$$
I know that they are set up like $sec(y)=x$ ...
$$sec(y)=frac{-2}{sqrt{3}}$$
$$sec(y)=-sqrt{2}$$
I got $frac{5pi}{6}$ for the first one and $frac{3pi}{4}$ for the second, but these are wrong and I'm not sure why.
The only feedback I got was "An inverse trigonometric function takes a numeric value as input, and returns an angle as the output."
calculus trigonometry inverse
edited Jan 30 at 19:13
Blue
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
$endgroup$
add a comment |
$begingroup$
$$sec^{-1}left(frac{-2}{sqrt{3}}right)$$
$$sec^{-1}left(-sqrt{2}right)$$
I know that they are set up like $sec(y)=x$ ...
$$sec(y)=frac{-2}{sqrt{3}}$$
$$sec(y)=-sqrt{2}$$
I got $frac{5pi}{6}$ for the first one and $frac{3pi}{4}$ for the second, but these are wrong and I'm not sure why.
The only feedback I got was "An inverse trigonometric function takes a numeric value as input, and returns an angle as the output."
calculus trigonometry inverse
edited Jan 30 at 19:13
Blue
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
$endgroup$
4
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
1
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
2
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39
add a comment |
$begingroup$
$$sec^{-1}left(frac{-2}{sqrt{3}}right)$$
$$sec^{-1}left(-sqrt{2}right)$$
I know that they are set up like $sec(y)=x$ ...
$$sec(y)=frac{-2}{sqrt{3}}$$
$$sec(y)=-sqrt{2}$$
I got $frac{5pi}{6}$ for the first one and $frac{3pi}{4}$ for the second, but these are wrong and I'm not sure why.
The only feedback I got was "An inverse trigonometric function takes a numeric value as input, and returns an angle as the output."
calculus trigonometry inverse
edited Jan 30 at 19:13
Blue
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
$endgroup$
$$sec^{-1}left(frac{-2}{sqrt{3}}right)$$
$$sec^{-1}left(-sqrt{2}right)$$
I know that they are set up like $sec(y)=x$ ...
$$sec(y)=frac{-2}{sqrt{3}}$$
$$sec(y)=-sqrt{2}$$
I got $frac{5pi}{6}$ for the first one and $frac{3pi}{4}$ for the second, but these are wrong and I'm not sure why.
The only feedback I got was "An inverse trigonometric function takes a numeric value as input, and returns an angle as the output."
calculus trigonometry inverse
calculus trigonometry inverse
edited Jan 30 at 19:13
Blue
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
edited Jan 30 at 19:13
Blue
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
edited Jan 30 at 19:13
Blue
49.4k870157
edited Jan 30 at 19:13
Blue
49.4k870157
edited Jan 30 at 19:13
Blue
49.4k870157
49.4k870157
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
asked Jan 30 at 19:01
SomeGuy2312SomeGuy2312
224
224
4
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
1
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
2
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39
add a comment |
4
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
1
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
2
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39
4
4
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
1
1
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
2
2
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Perhaps the answer is looking for degrees in which case, you have 150$^circ$ for the first one and 135$^circ$ for the second one.
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
add a comment |
$begingroup$
The restricted domain for $sec$ is not consistently defined. Some texts use;
$$y=sec^{-1} x iff sec y = x mbox{ and } yin [0,pi/2)cup [pi, 3pi/2). $$
So perhaps that is your snag. The reason for choosing this domain is that it makes the derivative formula nicer.
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
$endgroup$
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Perhaps the answer is looking for degrees in which case, you have 150$^circ$ for the first one and 135$^circ$ for the second one.
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
add a comment |
$begingroup$
Perhaps the answer is looking for degrees in which case, you have 150$^circ$ for the first one and 135$^circ$ for the second one.
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
$endgroup$
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
add a comment |
$begingroup$
Perhaps the answer is looking for degrees in which case, you have 150$^circ$ for the first one and 135$^circ$ for the second one.
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
$endgroup$
Perhaps the answer is looking for degrees in which case, you have 150$^circ$ for the first one and 135$^circ$ for the second one.
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
edited Jan 30 at 19:18
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
answered Jan 30 at 19:10
Akash PatelAkash Patel
168110
168110
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
add a comment |
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
1
1
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
$begingroup$
That's right, thanks Blue, just made the edit.
$endgroup$
– Akash Patel
Jan 30 at 19:18
add a comment |
$begingroup$
The restricted domain for $sec$ is not consistently defined. Some texts use;
$$y=sec^{-1} x iff sec y = x mbox{ and } yin [0,pi/2)cup [pi, 3pi/2). $$
So perhaps that is your snag. The reason for choosing this domain is that it makes the derivative formula nicer.
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
$endgroup$
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
|
show 2 more comments
$begingroup$
The restricted domain for $sec$ is not consistently defined. Some texts use;
$$y=sec^{-1} x iff sec y = x mbox{ and } yin [0,pi/2)cup [pi, 3pi/2). $$
So perhaps that is your snag. The reason for choosing this domain is that it makes the derivative formula nicer.
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
$endgroup$
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
|
show 2 more comments
$begingroup$
The restricted domain for $sec$ is not consistently defined. Some texts use;
$$y=sec^{-1} x iff sec y = x mbox{ and } yin [0,pi/2)cup [pi, 3pi/2). $$
So perhaps that is your snag. The reason for choosing this domain is that it makes the derivative formula nicer.
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
$endgroup$
The restricted domain for $sec$ is not consistently defined. Some texts use;
$$y=sec^{-1} x iff sec y = x mbox{ and } yin [0,pi/2)cup [pi, 3pi/2). $$
So perhaps that is your snag. The reason for choosing this domain is that it makes the derivative formula nicer.
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
answered Jan 30 at 19:30
B. GoddardB. Goddard
20.1k21543
20.1k21543
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
|
show 2 more comments
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
Yikes! A derivative formula for a discontinuous function!
$endgroup$
– Oscar Lanzi
Jan 30 at 20:28
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
@OscarLanzi It's not discontinuous everywhere. If you use the OP's domain, the derivative has an absolute value in it. If you use mine, you don't.
$endgroup$
– B. Goddard
Jan 30 at 20:31
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
Still discontinuous at a point. We would have to exclude the point of discontinuity from the derivative.
$endgroup$
– Oscar Lanzi
Jan 30 at 20:51
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
@OscarLanzi You have to do that with either domain. Heck, you have to do it with $tan$ and $1/x$. What's the big deal?
$endgroup$
– B. Goddard
Jan 30 at 21:00
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
$begingroup$
Derivatives of discontinuous functions have always made me lose it since high school :-S .
$endgroup$
– Oscar Lanzi
Jan 30 at 21:02
|
show 2 more comments
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4
$begingroup$
Your answers are correct.
$endgroup$
– KM101
Jan 30 at 19:06
1
$begingroup$
I agree that your answers are correct. Perhaps the source of the feedback has a bias for degrees in representing angles.
$endgroup$
– Blue
Jan 30 at 19:08
$begingroup$
Please show us your definition of ${rm sec}^{-1}$.
$endgroup$
– Christian Blatter
Jan 30 at 19:28
2
$begingroup$
Alternatively, @Blue, perhaps the arbiter of answers wanted the word “radians” after the numerical value.
$endgroup$
– Lubin
Jan 30 at 19:39