Solve for y: -sin^-1 y=sin^-1 x+C

Question

idealist10 · Answer

@ganeshie8

idealist10 · Answer

It's sine inverse.

ganeshie8 · Answer

what exactly do u feel is difficult here ?

idealist10 · Answer

The negative sign in sin^-1 y.

ganeshie8 · Answer

multiply -1 both sides, it goes away eh ?

idealist10 · Answer

So I got sin^-1 y=-sin^-1 x-C

ganeshie8 · Answer

$$\large  -\sin^{-1} y=\sin^{-1} x+C$$

$$\large  \sin^{-1} y=-\sin^{-1} x-C$$

ganeshie8 · Answer

yes, take "sin" both sides

ganeshie8 · Answer

$$\large  -\sin^{-1} y=\sin^{-1} x+C$$

$$\large  \sin^{-1} y=-\sin^{-1} x-C$$

$$\large  \sin(\sin^{-1} y)=\sin(-\sin^{-1} x-C)$$

idealist10 · Answer

I got y=-sin(sin^-1 x+C)

ganeshie8 · Answer

left side,  $\sin$ and $\sin^{-1}$ eat eachother out leaving you :

$$\large  y =\sin(-\sin^{-1} x-C)$$

ganeshie8 · Answer

yes, that looks good since "sin" is an odd function, so you can pull out the negative sign :

sin(-x) =  -sin(x)

idealist10 · Answer

So if I simplify more, I got y=-x-sin*C=-x-c since sine gets absorbed into C, the constant. Am I right?

ganeshie8 · Answer

NOPE.

sin(A+B) $\large \color{red}{\ne}$ sinA + sinB

ganeshie8 · Answer

$$\large  y =-\sin(\sin^{-1} x+C)$$

is the final simplified form, you're done !

idealist10 · Answer

But I have to use the identity sin(A-B)=sinAcosB-cosAsinB.

idealist10 · Answer

Thanks.