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Simply Trig Expression

by davemoosehead
Tags: expression, simply, trig
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davemoosehead
#1
Sep25-07, 09:02 PM
P: 26
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
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Avodyne
#2
Sep25-07, 10:03 PM
Sci Advisor
P: 1,229
Do you know a "double angle formula" that expresses cos(2t) in terms of sin(t)?
davemoosehead
#3
Sep25-07, 11:56 PM
P: 26
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
#4
Sep26-07, 01:04 AM
Sci Advisor
P: 1,229
Simply Trig Expression

Quote Quote by davemoosehead View Post
cos 2t = 1 - 2sint

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

cos 2t = 1 - 2(5x)
cos t = ( 1 - 2(5x) ) / (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
#5
Sep26-07, 01:16 AM
P: 26
sin²t = 5x²?
davemoosehead
#6
Sep26-07, 01:23 AM
P: 26
is cos 2t = 1 - 2(5x²) in simplest form? is that the answer?
Avodyne
#7
Sep26-07, 01:35 AM
Sci Advisor
P: 1,229
sin t = 5x, so sin2t=(5x)2=25x2.

But I'm sure you knew that ...
davemoosehead
#8
Sep26-07, 01:50 AM
P: 26
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
#9
Sep26-07, 12:10 PM
Sci Advisor
P: 1,229
That's it. You might want to simplify it further to 1-50x2, but that's a minor detail.


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