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Simply Trig Expression 
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#1
Sep2507, 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 


#2
Sep2507, 10:03 PM

Sci Advisor
P: 1,192

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



#3
Sep2507, 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? 


#4
Sep2607, 01:04 AM

Sci Advisor
P: 1,192

Simply Trig Expression



#5
Sep2607, 01:16 AM

P: 26

sin²t = 5x²?



#6
Sep2607, 01:23 AM

P: 26

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



#7
Sep2607, 01:35 AM

Sci Advisor
P: 1,192

sin t = 5x, so sin^{2}t=(5x)^{2}=25x^{2}.
But I'm sure you knew that ... 


#8
Sep2607, 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) 


#9
Sep2607, 12:10 PM

Sci Advisor
P: 1,192

That's it. You might want to simplify it further to 150x^{2}, but that's a minor detail.



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