1. ## 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)

2. ## Re: To solve K

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)

What are you actually trying to do with this question?

Edit: Didn't read the title. Just divide both sides by \begin{align*} \sin{\left( A + B \right)} \end{align*}.

3. ## Re: To solve K

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