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!
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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 $\displaystyle \begin{align*} \sin{\left( A + B \right)} \end{align*}$.
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
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