Question

How do you simplify and find the domain of

sin2(sin2x)

Answer 1

Do you need to simplify it? Or just find the domain?

Answer 2

If f(x)=sinx, what you have is

f(2f(2x))

If you know the domain of y=sin(x), then you know the domain of y=sin(2x).

Now think about the range of y=sin(2x).

Then the domain of y=sin[2(sin(2x))] will just take every element in the range of sin(2x) and multiply it by 2.

Answer 3

I have to simplify as well

Answer 4

Well then I would use the identity

sin(2x)=2sinx*cosx

Answer 5

ok, how do you simplify 2sin(sin2x)cos(2x)? It is the sin(sin2x) portion that I am struggling with.

Answer 6

Hmmm... ok, at first I didn't understand how you got that, but now I see what you did. Here is what I was thinking:

You have: sin2(sin2x)

sin(2x)=2sinxcosx so plug that in. you get:

sin[2(2sinxcosx)]=sin(4sinxcosx)

Answer 7

What I said above about the domain is wrong. Duh. I'm sorry....

Remember that for a composition, x is in the domain of the composition if:

1. x is in the domain of the "inner function" (here that is sin(2x), so domain of that is all reals

and

2. the output of the "inner function" is in the domain of the "outer function" (here the "outer function" is also sin(2x) which has domain all reals, so EVERYTHING that you can output from the inner function is in the domain of the outer function.)

Hence, the whole domain of the inner function is the domain of the composition.