For $0 < x < 1$, express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$
up vote
2
down vote
favorite
I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.
I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.
For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
That's where I'm stuck.
algebra-precalculus trigonometry
edited Nov 23 at 19:35
Robert Howard
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
add a comment |
up vote
2
down vote
favorite
I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.
I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.
For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
That's where I'm stuck.
algebra-precalculus trigonometry
edited Nov 23 at 19:35
Robert Howard
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
2
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
3
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.
I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.
For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
That's where I'm stuck.
algebra-precalculus trigonometry
edited Nov 23 at 19:35
Robert Howard
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
I need to express $sin[sin^{-1}(x) + cos^{-1}(x)]$, in terms of $x$, if $0 < x < 1$.
I'm not sure how to solve this. If I knew the value of $x$, I would try and apply the identity, $sin(A-B)=sin(A)cos(B)+ cos(A)sin(A)$, but since the answer is the real number $1$, I don't see how that would work.
For example:
$$sin[sin^{-1}(x) + cos^{-1}(x)]$$
$$sin[sin^{-1}(x)] cdot cos[cos^{-1}(x)] + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x cdot x + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
$$x^2 + cos[sin^{-1}(x)] cdot sin[cos^{-1}(x)]$$
That's where I'm stuck.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 23 at 19:35
Robert Howard
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
edited Nov 23 at 19:35
Robert Howard
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
edited Nov 23 at 19:35
Robert Howard
1,9181822
edited Nov 23 at 19:35
Robert Howard
1,9181822
edited Nov 23 at 19:35
Robert Howard
1,9181822
1,9181822
asked Nov 23 at 19:09
LuminousNutria
1709
asked Nov 23 at 19:09
LuminousNutria
1709
asked Nov 23 at 19:09
LuminousNutria
1709
1709
2
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
3
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12
add a comment |
2
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
3
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12
2
2
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
3
3
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
|
show 3 more comments
up vote
1
down vote
Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.
Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.
edited Nov 23 at 19:57
Timothy Cho
789519
answered Nov 23 at 19:19
Alex
111
add a comment |
up vote
0
down vote
Hint: Use
$$arcsin x = y_1 implies sin y_1 = x$$
$$arccos x = y_2 implies cos y_2 = x$$
and
$$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$
to get
$$implies y_2 = frac{pi}{2}-y_1$$
So what does $y_1+y_2$ become?
answered Nov 23 at 19:20
KM101
3,892417
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
|
show 2 more comments
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
|
show 3 more comments
up vote
2
down vote
accepted
Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
|
show 3 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
Using the fact $$cos(u)=sqrt{1-sin^2 u}\sin(u)=sqrt{1-cos^2 u}$$we have $$cos [sin^{-1}x]cdot sin [cos^{-1}x]=sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)}=1-x^2$$therefore $$large sin[sin^{-1}x+cos^{-1}x]=1$$
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
edited Nov 23 at 19:35
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
answered Nov 23 at 19:13
Mostafa Ayaz
13.6k3836
13.6k3836
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
|
show 3 more comments
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
I'm sorry, I don't understand how that helps.
– LuminousNutria
Nov 23 at 19:28
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
Then you can substitute $u=sin^{-1}x$ and $u=cos^{-1}x$ in the first and second equations respectively
– Mostafa Ayaz
Nov 23 at 19:29
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
So, instead of $sin[sin^{-1}(x) + cos^{-1}(x)]$, I'll have $sin[u + u]$?
– LuminousNutria
Nov 23 at 19:32
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
I clarified my answer a bit more. Hope it help now!
– Mostafa Ayaz
Nov 23 at 19:36
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
So, why does $sqrt{1-sin^2 (sin^{-1}x)}cdot sqrt{1-cos^2 (cos^{-1}x)} = 1 - x^2$ ?
– LuminousNutria
Nov 23 at 19:41
|
show 3 more comments
up vote
1
down vote
Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.
Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.
edited Nov 23 at 19:57
Timothy Cho
789519
answered Nov 23 at 19:19
Alex
111
add a comment |
up vote
1
down vote
Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.
Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.
edited Nov 23 at 19:57
Timothy Cho
789519
answered Nov 23 at 19:19
Alex
111
add a comment |
up vote
1
down vote
up vote
1
down vote
Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.
Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.
edited Nov 23 at 19:57
Timothy Cho
789519
answered Nov 23 at 19:19
Alex
111
Write $cos(sin^{-1} x)$ as $sin(fracpi2 - sin^{-1}x)$ and expand using the identity.
Similarly, solve the other term by writing $sin(cos^{-1} x)$ as $cos(fracpi2 - cos^{-1} x)$ and expanding it using an identity.
edited Nov 23 at 19:57
Timothy Cho
789519
answered Nov 23 at 19:19
Alex
111
edited Nov 23 at 19:57
Timothy Cho
789519
edited Nov 23 at 19:57
Timothy Cho
789519
edited Nov 23 at 19:57
Timothy Cho
789519
789519
answered Nov 23 at 19:19
Alex
111
answered Nov 23 at 19:19
Alex
111
answered Nov 23 at 19:19
Alex
111
111
add a comment |
add a comment |
up vote
0
down vote
Hint: Use
$$arcsin x = y_1 implies sin y_1 = x$$
$$arccos x = y_2 implies cos y_2 = x$$
and
$$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$
to get
$$implies y_2 = frac{pi}{2}-y_1$$
So what does $y_1+y_2$ become?
answered Nov 23 at 19:20
KM101
3,892417
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
|
show 2 more comments
up vote
0
down vote
Hint: Use
$$arcsin x = y_1 implies sin y_1 = x$$
$$arccos x = y_2 implies cos y_2 = x$$
and
$$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$
to get
$$implies y_2 = frac{pi}{2}-y_1$$
So what does $y_1+y_2$ become?
answered Nov 23 at 19:20
KM101
3,892417
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
|
show 2 more comments
up vote
0
down vote
up vote
0
down vote
Hint: Use
$$arcsin x = y_1 implies sin y_1 = x$$
$$arccos x = y_2 implies cos y_2 = x$$
and
$$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$
to get
$$implies y_2 = frac{pi}{2}-y_1$$
So what does $y_1+y_2$ become?
answered Nov 23 at 19:20
KM101
3,892417
Hint: Use
$$arcsin x = y_1 implies sin y_1 = x$$
$$arccos x = y_2 implies cos y_2 = x$$
and
$$cos theta = sin bigg(frac{pi}{2}-thetabigg)$$
to get
$$implies y_2 = frac{pi}{2}-y_1$$
So what does $y_1+y_2$ become?
answered Nov 23 at 19:20
KM101
3,892417
edited Nov 23 at 19:24
answered Nov 23 at 19:20
KM101
3,892417
answered Nov 23 at 19:20
KM101
3,892417
answered Nov 23 at 19:20
KM101
3,892417
3,892417
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
|
show 2 more comments
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
You've got to be careful here $sin y_1=cos y_2$ doesn't necessarily imply that $y_2=frac{pi}{2}-y_1$. For example, $sin (frac{pi}{2})=cos (2pi)$.
– Anurag A
Nov 23 at 19:24
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
But $2pi = 0 $ (in radians), so you get $sin frac{pi}{2} = cos 0$.
– KM101
Nov 23 at 19:27
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
what do you mean by $2pi=0$? !!!
– Anurag A
Nov 23 at 19:30
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
(Awful wording, I edited the comment.) I meant $0$ radians and $2pi$ radians are the same.
– KM101
Nov 23 at 19:32
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
No they are not. $0$ radian is $0^{circ}$ and $2pi$ radian is $360^{circ}$.
– Anurag A
Nov 23 at 19:34
|
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2
socratic.org/questions/how-do-you-prove-arcsin-x-arccos-x-pi-2
– lab bhattacharjee
Nov 23 at 19:12
3
Hint: $arcsin x+ arccos x=frac{pi}{2}$.
– Anurag A
Nov 23 at 19:12