Trig Product: Simplifying sina+sinb+sin(a+b) as a Product

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SUMMARY

The discussion focuses on simplifying the expression \( \sin a + \sin b + \sin(a+b) \) into a product form. The key transformation involves expressing \( \sin(a+b) \) as \( 2 \sin((a+b)/2) \cos((a-b)/2) \) and factoring it further. The final result combines terms to yield a product involving \( \sin((a+b)/2) \) and \( \cos(a/2) \cos(b/2) \). This method efficiently consolidates the trigonometric identities into a simplified product.

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  • Understanding of trigonometric identities
  • Familiarity with sine and cosine functions
  • Knowledge of factorization techniques in trigonometry
  • Basic algebraic manipulation skills
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  • Study the derivation of the sine addition formula
  • Learn about product-to-sum identities in trigonometry
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Students and educators in mathematics, particularly those focusing on trigonometry, as well as anyone looking to enhance their skills in simplifying trigonometric expressions.

tatoo5ma
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Hello,
I have to write {\it sina}+{\it sinb}+\sin \left( a+b \right) as a product.
Here is what I began with, then I got stuck...
2\,\sin \left( (a+b)/2 \right) \cos \left( (a-b)/2 \right) +{\it sina}\,{\it cosb}+{\it cosa}\,{\it sinb}

Anyone please can help me out, that would be great, thank you! :smile:
 
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tatoo5ma, Instead of expanding sin(a+b), what else can you do? If you write it as a trignometric function of (a+b)/2 can you see what happens?
 
Last edited:
sure

Sin (a+b) = Sin (2(a+b)/2) = 2 cos((a+b)/2)Sin((a+b)/2)
 
yeah, then u can factor with Sin((a+b)/2) and then make cos((a-b)/2)+cos((a+b)/2) equal to 2cos(a/2)cos(b/2) and u have everythin factorized :smile:
 
Last edited:

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