If #sinA*sin(B-c)=sinC*sin(A-B)# then show that # a^2,b^2,c^2# are in AP.?

Question

Nolan Alexander · Accepted Answer

#sinA*sin(B-C)=sinC*sin(A-B)# #=>2sin(pi-(B+C))*sin(B-C)=2sin(pi-(A+B))*sin(A-B)# #=>2sin(B+C)*sin(B-C)=2sin(A+B)*sin(A-B)# #=>cos(2C)-cos(2B)=cos(2B)-cos(2A)# #=>1-2sin^2(C)-1+2sin^2(B)=1-2sin^2(B)-1+2sin^2(A)# #=>sin^2(B)-sin^2(C)=sin^2(A)-sin^2(B)# #=>4R^2sin^2(B)-4R^2sin^2(C)=4R^2sin^2(A)-4R^2sin^2(B)# #=>b^2-c^2=a^2-b^2# This proves that # a^2,b^2,c^2# are in AP

Adam Aguirre · Answer

undefined Given the equation sinA*sin(B-c)=sinC*sin(A-B), let's start by expanding both sides using the trigonometric identity: sin(x - y) = sinx*cosy - cosx*siny.

Expanding the left-hand side (LHS):
sinA*sin(B - c) = sinA*(sinB*cosc - cosB*sinc)
= sinA*sinB*cosc - sinA*cosB*sinc

Expanding the right-hand side (RHS):
sinC*sin(A - B) = sinC*(sinA*cosB - cosA*sinB)
= sinC*sinA*cosB - sinC*cosA*sinB

Now, equating LHS and RHS:
sinA*sinB*cosc - sinA*cosB*sinc = sinC*sinA*cosB - sinC*cosA*sinB

sinA*sinB*cosc - sinC*sinA*cosB = sinA*cosB*sinc - sinC*cosA*sinB

sinA*sinB*cosc - sinC*sinA*cosB = sinA*cosB*(sinc - sinC)

Dividing both sides by sinA*cosB:
cosc - sinC = sinc - sinC

c - C = c - C

Since both sides are equal, the equation is satisfied.

Now, we need to prove that a^2, b^2, and c^2 are in arithmetic progression (AP).
We know that:
a = 2R*sinA
b = 2R*sinB
c = 2R*sinC

where R is the radius of the circumcircle of triangle ABC.

Now, squaring each of these and rearranging:
a^2 = 4R^2*sin^2A
b^2 = 4R^2*sin^2B
c^2 = 4R^2*sin^2C

Now, it's evident that:
b^2 - a^2 = 4R^2*(sin^2B - sin^2A)
c^2 - b^2 = 4R^2*(sin^2C - sin^2B)

Given sinA*sin(B-c)=sinC*sin(A-B), we can conclude that sin^2A = sin^2C. Hence, sin^2B - sin^2A = 0 and sin^2C - sin^2B = 0.

Therefore, b^2 - a^2 = 0 and c^2 - b^2 = 0, implying that a^2, b^2, and c^2 are in arithmetic progression.