Trigonometric Prove for Triangle ABC

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
 
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

Thank you. Do you enjoy David Cohen questions?
 


Write your reply...

Members online

No members online now.

Trending content

Forum statistics

Threads
2,527
Messages
9,856
Members
696
Latest member
fairdistribution
Back
Top