Thread: To solve K

i was given this qn.... sin(A-B) = k sin(A+B)

sin(A-B) = sinAcosB + sinBcosA
= 1/2[sin(A+B)+sin(A-B)] - 1/2[sin(A+B)-sin(A-B)]

im not sure if im on the right track. cant seem to continue cos i will always return to sin(A-B)

PLEASE HELP!

Originally Posted by ZGOON

i was given this qn.... sin(A-B) = k sin(A+B)

sin(A-B) = sinAcosB + sinBcosA
= 1/2[sin(A+B)+sin(A-B)] - 1/2[sin(A+B)-sin(A-B)]

im not sure if im on the right track. cant seem to continue cos i will always return to sin(A-B)

PLEASE HELP!

What are you actually trying to do with this question?

Edit: Didn't read the title. Just divide both sides by .

cos i need to prove that sin(A-B) = k sin(A+B)


then from there i need to prove


tanA = [(1+k)/(1-k)] tanB