Question

how to solve this

Answer 1

Sorry, could you rewrite or draw the question as its a little unclear.
Thankx

Answer 2

Is that the full question or does it equal to something ? Also, does it give you the value of x at which to evaluate?

Answer 3

did u mean simplify?

Answer 4

yes
@sauravshakya

Answer 5

\[\sqrt{\cos ^{-1}(sinx)} = (\cos ^{-1}(sinx))^{\frac{ 1 }{ 2 }}\]
\[=\cos ^{-1}(\sin x)^{\frac{ 1 }{ 2 }}\]
this is how far i could go on with the simplification....

Answer 6

@kausarsalley u r right
i have also done the same produre
as this sum

Answer 7

* for this sum

Answer 8

@msingh do you mean procedure??

Answer 9

yes

Answer 10

\[\sin(x)=\cos(\frac{ \pi }{ 2 }-x)\rightarrow \sqrt{\cos^{-1}\left( \sin(x) \right)}=\sqrt{\cos^{−1}\left( \cos \left( \frac{ \pi }{ 2}-x\right) \right)}\] and that equals\[\sqrt{\frac{ \pi }{ 2 }-x}\]

Answer 11

let y= (cos^-1(sinx))^1/2
y^2=cos^-1(sinx)
cosy^2=sinx ......i
cos^2(y^2)=sin^2x
1-sin^2(y^2)=sin^2x
1-sin^2x=sin^2(y^2)
cos^2x=sin^2(y^2)
sin(y^2)=cosx ....ii
Now, from i and ii, sin(2y^2)=sin(2x)
y=x^(1/2)
Find the mistake...

Answer 12

gud
but i didn't find any mistake

Answer 13

@sauravshakya
would u pse explain me step after (i) this

Answer 14

squaring both sides

Answer 15

and use sin^2(x)+cos^2(x)=1

Answer 16

but how u get this
cos^2(y^2)=sin^2x

Answer 17

y= (cos^-1(sinx))^1/2
y^2=cos^-1(sinx)
cosy^2=sinx
Now square both sides

Answer 18

k

Answer 19

There is a mistake... any one got it?

Answer 20

@sauravshakya
if
cosy^2=sinx ......i

sin(y^2)=cosx ....ii
then how you got this

Now, from i and ii, sin(2y^2)=sin(2x)
y=x^(1/2)

Answer 21

i got it

Answer 22

u got the mistake?