How to simplify an iterated trigonometric expression

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Simplifying the expression cos(sin x) is challenging, as it typically cannot be simplified beyond its current form. The discussion highlights that one approach involves using the partial sums of the Taylor series for these functions. Additionally, the Fourier series of cos(sin x) may relate to Bessel functions, which could provide further insights. Alternative identities, such as cos²(sin x) + sin²(sin x) = 1, were suggested but do not lead to significant simplification. Overall, the consensus is that substantial simplification of cos(sin x) is unlikely.
Leo Liu
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Homework Statement
.
Relevant Equations
.
eg ##\cos (\sin x)##
Asking this question out of curiosity.
 
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Leo Liu said:
Homework Statement:: .
Relevant Equations:: .

eg ##\cos (\sin x)##
Asking this question out of curiosity.
I don't see how you're going to be able to get anything simpler than that.
 
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Typically this can only be done with the partial sums of the Taylor series of the functions. There are a variety of ways to calculate a partial sum of the composition, including matrix multiplication.
 
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If you are interested about the Fourier series of cos(cos x) or cos (sin x) I think they are related to the Bessel functions.
 
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Delta2 said:
If you are interested about the Fourier series of cos(cos x) or cos (sin x) I think they are related to the Bessel functions.
Thank you this is the best answer I got :D
 
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Or maybe you want something like :

##Cos^2(sinx)+ Sin^2(sinx)=1##

So that ##Cos(sinx)=\sqrt {1-Sin^2(sinx)}##?
 
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WWGD said:
Or maybe you want something like :

##Cos^2(sinx)+ Sin^2(sinx)=1##

So that ##Cos(sinx)=\sqrt {1-Sin^2(sinx)}##?
I thought of something like this, as well as a Taylor or Maclaurin series, but none of these seemed like they would serve to simplify the given expression.
 
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