Question

in triangle ABC,sigma sin(A-B)/sinAsinB
​

Answer 1

Step-by-step explanation:

In △ABC,A+B+C=π

Given sinA.sin(B−C)=sinCsin(A−B)

⟹sin(π−(B+C))sin(B−C)=sin(π−(A+B))sin(A−B)

⟹sin(B+C)sin(B−C)=sin(A+B)sin(A−B)

⟹sin2B−sin2C=sin2A−sin2B

⟹2sin2B−2sin2C=2sin2A−2sin2B

⟹(1−cos2B)−(1−cos2C)=(1−cos2A)−(1−cos2B)

⟹cos2C−cos2B=cos2B−cos2A

⟹2cos2B=cos2A+cos2C

So cos2A,cos2B,cos2C are in A.P