Trigonometric Prove for Triangle ABC

[nycmathguy]

David Cohen Just for you. Enjoy.

 

[MathLover1]

prove that sin(A-B)/sin(A+B)=(a^2+b^2)/c^2

use identities:
sin(A-B)=sin(A) cos(B) - cos(A) sin(B)
sin(A+B)=sin(A) cos(B) + cos(A) sin(B)

sin(A-B)/sin(A+B)=(sin(A) cos(B) - cos(A) sin(B))/(sin(A) cos(B) + cos(A) sin(B))

sin(A-B)/sin(A+B)=(2ac*cos(B)-2bc*cos(A))/(2ac*cos(B)+2bc*cos(A))

sin(A-B)/sin(A+B)=(a^2-b^2+c^2-(-a^2+b^2+c^2))/((a^2-b^2+c^2)+(-a^2+b^2+c^2))

sin(A-B)/sin(A+B)=(a^2-b^2+c^2+a^2-b^2-c^2)/(a^2-b^2+c^2-a^2+b^2+c^2)

sin(A-B)/sin(A+B)=(2a^2-2b^2)/(2c^2)

sin(A-B)/sin(A+B)=(a^2-b^2)/c^2

 

[nycmathguy]

prove that sin(A-B)/sin(A+B)=(a^2+b^2)/c^2

use identities:
sin(A-B)=sin(A) cos(B) - cos(A) sin(B)
sin(A+B)=sin(A) cos(B) + cos(A) sin(B)

sin(A-B)/sin(A+B)=(sin(A) cos(B) - cos(A) sin(B))/(sin(A) cos(B) + cos(A) sin(B))

sin(A-B)/sin(A+B)=(2ac*cos(B)-2bc*cos(A))/(2ac*cos(B)+2bc*cos(A))

sin(A-B)/sin(A+B)=(a^2-b^2+c^2-(-a^2+b^2+c^2))/((a^2-b^2+c^2)+(-a^2+b^2+c^2))

sin(A-B)/sin(A+B)=(a^2-b^2+c^2+a^2-b^2-c^2)/(a^2-b^2+c^2-a^2+b^2+c^2)

sin(A-B)/sin(A+B)=(2a^2-2b^2)/(2c^2)

sin(A-B)/sin(A+B)=(a^2-b^2)/c^2

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