Inversing an equation with sin in it?

[ram012593]

My question is does (y=(-1/4)sin(x/4))^-1 equal this y = (4)sin^-1(4x) ? If so can this be further simplified or am I doing it totally wrong? I apologize if this isn't posted in the right section or if I put too many parenthesis I wanted to make sure what I am asking was clear as possible.

Thanks much in Advance!!!!!

 

[Deleted member 4993]

My question is does (y=(-1/4)sin(x/4))^-1 equal this y = (4)sin^-1(4x) .... Incorrect? If so can this be further simplified or am I doing it totally wrong? I apologize if this isn't posted in the right section or if I put too many parenthesis I wanted to make sure what I am asking was clear as possible.

Thanks much in Advance!!!!!

y = -1/4 * sin(x/4)

exchange x ↔ y

x = -1/4 * sin(y/4)

Solve for y

-4x = sin(y/4)

Now continue.....

 

[HallsofIvy]

You do need to know: because sin(x) is an odd function (sin(-x)= -sin(x)), so is \(\displaystyle sin^{-1}(x)\): \(\displaystyle sin^{-1}(-4x)= -sin^{-1}(4x)\).

(You cannot, of course, take the "4" out of the arcsine that way!)

 

[ram012593]

Originally Posted by Subhotosh Khan

y = -1/4 * sin(x/4)

exchange x ↔ y

x = -1/4 * sin(y/4)

Solve for y

-4x = sin(y/4)

Now continue.....

Actually that is exactly the method I tried to use but you actually stopped exactly where I probably screwed up you see I don't know what the inverse operation of sin is so I would be unable to continue from there I figured it would just be sin^-1 but I guess not. These: are the steps I took:

x=(-1/4)sin(x/4) swapped x and y

-4x=sin(y/4) multiplied both sides by -4 (I noticed I forgot the negative in the original)

sin^-1(-4x)=y/4 did what I thought was the inverse operation of sin to both sides and canceled the one on the right (probably where I messed up mostly)

4(sin^-1(-4x)) = y multiplied both sides by 4

Thanks for all the replies!!

 

[Deleted member 4993]

Actually that is exactly the method I tried to use but you actually stopped exactly where I probably screwed up you see I don't know what the inverse operation of sin is so I would be unable to continue from there I figured it would just be sin^-1 but I guess not. These: are the steps I took:

x=(-1/4)sin(y/4) swapped x and y

-4x=sin(y/4) multiplied both sides by -4 (I noticed I forgot the negative in the original)

sin^-1(-4x)=y/4 did what I thought was the inverse operation of sin to both sides and canceled the one on the right (probably where I messed up mostly)

4(sin^-1(-4x)) = y multiplied both sides by 4

since sin(-Θ) = -sin(Θ)

y = -4sin^-1(4x) ........................................Inverse function

Thanks for all the replies!!

Notice the corrections above.

check: we know f[f^-1(x)] = x then

f(x) =(-1/4)sin(x/4)

f^-1(x) = -4sin^-1(4x)

f[f^-1(x)] =(-1/4)sin{[f^-1(x)]/4} = (-1/4)sin{[-4sin^-1(4x)]/4)}= (1/4)sin{[sin^-1(4x)])} = 1/4 * (4x) = x

checks....