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Question
Hi,

I have a question regarding trigonometry.

Q: If A + B + C = 180deg, prove that sin A + sin B + sin C = 4 cos (A/2) cos (B/2) cos (C/2).

Solution:
LHS = (sin A + sin B) + sin (180deg - (A+B))
   = 2sin (A+B /2) cos (A-B /2) + sin (A+B)
   = 2sin (A+B /2) cos (A-B /2) + 2 sin (A+B /2) cos (A+B /2)
and the solution continues...

What I am pondering about is how sin (A+B) can be expressed as 2 sin (A+B /2) cos (A+B /2). Thanks!

Answer
sin(A+B) = 2sin(A+B/2)cos(A+B/2) ?

The only way I can reconcile this expression is to suggest that the argument A+B/2 should actually be (A+B)/2, in which case the famiiar identity

sin(2x) = 2(sinx)(cosx) works for x = (A+B)/2.

The other expressions in our partial solution are somewhat of a mystery to me.

Hope this helps.

Randy  

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randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

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