# Simply Trig Expression

1. Sep 25, 2007

1. The problem statement, all variables and given/known data

Simplify the expression:

cos(2sin^-1 (5x))

2. Relevant equations

Fundamental identities: 1 = sin^2 ϑ + cos^2 ϑ : I think you use this one?

3. The attempt at a solution

Let y=2sin^-1(5x)
sin y = 10x

so, you plug in?
1 = 10x^2 + cos^2 y

not really sure if im on the right path or what to do next

2. Sep 25, 2007

### Avodyne

Do you know a "double angle formula" that expresses cos(2t) in terms of sin(t)?

3. Sep 25, 2007

cos 2t = 1 - 2sin²t

let t = sin^-1 5x
so sin t = 5x

cos 2t = 1 - 2(5x²)
cos t = ( 1 - 2(5x²) ) / (2)

is this correct?

Last edited: Sep 26, 2007
4. Sep 26, 2007

### Avodyne

cos(2t)=1-2(5x2) is almost correct; if t=arcsin(5x), what is sin2t ? But your last line is not correct. (Can you see why?) However, you don't need the last line; you have simplified the expression, and you're done!

5. Sep 26, 2007

sin²t = 5x²?

6. Sep 26, 2007

is cos 2t = 1 - 2(5x²) in simplest form? is that the answer?

7. Sep 26, 2007

### Avodyne

sin t = 5x, so sin2t=(5x)2=25x2.

But I'm sure you knew that ...

8. Sep 26, 2007

Oh duh, ok from the start:

cos(2 arcsin 5x)
Let t = arcsin 5x
so, sin t = 5x

Since cos 2t = 1 - 2sin²t

cos 2t = 1 - 2(5x)²
cos 2t = 1 - 2(25x²)

9. Sep 26, 2007

### Avodyne

That's it. You might want to simplify it further to 1-50x2, but that's a minor detail.