Proving the Identity: sin2(x)-sin2(x)=sin(x+y)sin(x-y)?

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Homework Help Overview

The discussion revolves around proving a trigonometric identity involving sine functions, specifically the expression sin²(x) - sin²(y) = sin(x+y)sin(x-y).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove the identity but struggles to manipulate one side to match the other. Some participants suggest clarifying the expression to sin²(x) - sin²(y) and recommend using sum and difference formulas for sine.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the proof. There is acknowledgment of a correction regarding the initial expression, indicating a productive direction in the discussion.

Contextual Notes

There is a noted confusion regarding the initial expression, which was incorrectly stated as sin²(x) - sin²(x). The discussion highlights the importance of accurately defining terms in trigonometric identities.

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Homework Statement


Prove this is an identity:
sin2(x)-sin2(x)=sin(x+y)sin(x-y)


Homework Equations


N/A


The Attempt at a Solution


I have made a lot of attempts but can not get one side to equal the other. I know It's something really simple I am missing, but can't figure it out.
 
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Clearly, you meant sin(x)^2-sin(y)^2. Use the sum formula for sin(a+b) and sin(a-b) and expand it. Then use cos^2-1=sin^2. Expand again.
 
You are correct, I did mean sin2(x) - sin 2(y). And you clearly meant 1-cos2=sin2.
Thanks.
 
Right.
 

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