Question

Please Solve:-

Answer 1

\[\cos^{-1} (x)-\sin^{-1} (x)=0\]

Answer 2

Please solve for "x" :)

Answer 3

I have done as this:-
\[\cos^{-1} (x)-\sin^{-1} (x)+\sin^{-1} (x)=\sin^{-1} (x).\] \[\pi/2 = 2\sin^{-1} (x).\]

Answer 4

you can represent this in the following right angled triangle:

Answer 5

here:\[\cos(\theta)=\sin(\theta)=\frac{x}{1}=1\]so what is \(\theta\)?

Answer 6

sorry - that should be an x at the right hand side

Answer 7

\[\cos(\theta)=\sin(\theta)=\frac{x}{1}=x\]

Answer 8

\[\sin^{-1} (x)\] or \[\cos^{-1} (x)\]

Answer 9

the equations I wrote can be rewritten as:\[\theta=\cos^{-1}(x)=\sin^{-1}(x)\]

Answer 10

so theta should be pi/4

Answer 11

yes

Answer 12

& we get x=1/sqrt(2)

Answer 13

yes

Answer 14

but in my method I am getting 2 values for "x".\[(1/\sqrt{2}) ;-(1/\sqrt{2})\]

Answer 15

my method is stated above. please help:)

Answer 16

how did you get to:\[\pi/2 = 2\sin^{-1} (x)\]

Answer 17

since in my course ; we state a property as:-\[\sin^{-1} (\alpha)+\cos^{-1} (\alpha)=\pi/2 \] Where alpha is an angle:)

Answer 18

did u get that:)

Answer 19

but by using this substitution you are losing the restriction that:\[\cos^{-1}(x)=\sin^{-1}(x)\]

Answer 20

k! thanx I got it:)

Answer 21

so from your two solutions, you need to confirm which one satisfies the original problem

Answer 22

yw :)

Answer 23

:)