How do you calculate #sin(arcsin 2x)#?

1 Answer
Feb 8, 2015

It will be #2x#

#arcsin 2x# means find the angle of which the sine is #2x#

#sin# means find the sine.
So you first find the angle, and take the sine of it.

So the real question is:
Find the sine of the angle that 'belongs' to a sine of #2x#

In short
#sin# and #arcsin# are two opposing operations, they cancel each other out, just like square and square root.

There are limitations, since #sin# must be between #-1# and #+1#,
so #2x# is also limited to these boundaries:

#-1/2<=x<=+1/2#