Solve Trigonometry Proof: A+B+C=π

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To prove that asin(B-C) + bsin(C-A) + csin(A-B) = 0 given A + B + C = π, one approach is to express one variable in terms of the others and substitute it into the equation. Utilizing the sine identities, particularly those involving shifts by π, is crucial for simplifying the expressions. The relationship sin(x - π) = -sin(x) can be particularly useful in this proof. Additionally, referencing trigonometric identities can aid in the solution process. Overall, the key lies in strategic substitutions and applying the properties of sine functions effectively.
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Homework Statement



if A+B+C=π, prove that asin(B-C)+bsin(C-A)+csin(A-B)=0

Homework Equations



sin(x-y)=sinxcosy-cosxsiny

The Attempt at a Solution



I tried to expand but to no avail. Any help is appreciated.
 
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anthonych414 said:

Homework Statement



if A+B+C=π, prove that asin(B-C)+bsin(C-A)+csin(A-B)=0

Homework Equations



sin(x-y)=sinxcosy-cosxsiny

The Attempt at a Solution



I tried to expand but to no avail. Any help is appreciated.

Here are some advices:

→Take A + B + C = π and solve for one of the variables. Then, substitute C, B or A for asin(B-C)+bsin(C-A)+csin(A-B)=0.
→Since you get the shift by π in sine expressions, you need to use these formulas:

sin(x - π) = -sin(x) and sin(-x + π) = sin(x) [Make sure that when get the expression like sin(A + B - π), we have -sin(A + B)]

Oh! This link might help you! http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry.2C_shifts.2C_and_periodicity

Good luck, and let me know if you have comments or problems. :D
 
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