A Generalized trigonometric identity for Cos(x_1++x_n)?

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Hi Everyone,
Do there exist any explicit formula for Cos(x_1+x_2+...+x_n) as a sum of products of Sin(x_i) & Cos(x_i)? Or we need to expand using Cos(A+B), Sin(A+B) again & again?
If it exists then what is about Sin(x_1+x_2+...+x_n)?

[It is understood that there will be 2^(n-1) number of terms in the sum each of which will be
a product of n number of Sin(x_i) &/ Cos(x_i)]

Regards
Naturepaper
 
Mathematics news on Phys.org
http://en.wikipedia.org/wiki/Trigonometric_identity

That link has some things that are related, but not exactly the same.

I'm not sure if a formula can be found, but you can try it. Expand it for a couple of cases, try to find a formula and use induction.
 
Thank you Gib Z for your reply. But the link you given doesn't answer the question. It was for infinite sum whereas I need for finite terms. Anyway I'm waiting for more reply...Cheers
NaturePaper
 
NaturePaper, the same page gives
\sec(\theta_1 + \cdots + \theta_n) = \frac{\sec\theta_1 \cdots \sec\theta_n}{e_0 - e_2 + e_4 - \cdots}.
From that, it's obvious that
\cos(\theta_1 + \cdots + \theta_n) = \frac{e_0 - e_2 + e_4 - \cdots}{\sec\theta_1 \cdots \sec\theta_n}.
 
@adriank,
oh, very nice-I missed it. Thanks.

Intuitively, Sin(x_1+x_2+...+x_n)=sinx_1.sinx_2...sinx_n(e_1-e_3+e_4-...), where e_i are in terms of cotx_i. I'll check it.

Actually I need a formula to convert the product of K number of sinx_i and (n-k) number of cosx_i into sum of Sin and cosine terms.

If I get the expressions for Sin(x_1+x_2+...+x_n) and cos(x_1+x_2+...+x_n), after some manipulation I can get the required formula.

Thanks & Regards
NaturePaper
 
Got it.

Sin(x_1+x_2+...+x_n)=cosx_1cosx_2...cosx_n(e_1-e_3+e_4-...).

By the way, I'll post a possible proof and detailed formula soon.

cheers,
NaturePaper
 
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