How Do I Simplify this Trig Expression: cos(2sin^-1 (5x))?

[davemoosehead]

Homework Statement

Simplify the expression:

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

Homework Equations

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

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 I am on the right path or what to do next

 

[Avodyne]

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

 

[davemoosehead]

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?

 

[Avodyne]

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?

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!

 

[davemoosehead]

sin²t = 5x²?

 

[davemoosehead]

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

 

[Avodyne]

sin t = 5x, so sin²t=(5x)²=25x².

But I'm sure you knew that ...

 

[davemoosehead]

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)

 

[Avodyne]

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