Where does this trig identiy come from?

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Discussion Overview

The discussion revolves around the origin of a specific trigonometric identity involving cosine functions, specifically the identity: cos(x-a)cos(b-x) = \frac{1}{2}[cos(a+b-2x) + cos(a-b)]. Participants are seeking clarification on its derivation and validity.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses confusion about the identity's origin and notes difficulty finding it in standard references, only locating it on WolframAlpha.
  • Another participant suggests that the identity can be derived from the known sum of cosines identity: \cos \alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}, implying a connection but not providing a full derivation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the identity's derivation, and the discussion includes both confusion and a proposed connection to another known identity without full resolution.

Contextual Notes

There are limitations regarding the clarity of the derivation process and the reliance on established identities, which may not be universally recognized or accessible to all participants.

vincent_vega
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Our teacher used the following on a problem solution:

cos(x-a)cos(b-x) = \frac{1}{2}[cos(a+b-2x) + cos(a-b)]

Where does this come from? I can't find it in anywhere (except for wolframalpha). Thanks.
 
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thanks for the heads up
 
vincent_vega said:
Our teacher used the following on a problem solution:

cos(x-a)cos(b-x) = \frac{1}{2}[cos(a+b-2x) + cos(a-b)]

Where does this come from? I can't find it in anywhere (except for wolframalpha). Thanks.



It comes from the known identity for sum of cosines: \cos \alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}

DonAntonio
 

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