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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!

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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