2305sin2A-sin2B
=(sinA-sinB)(sinA+sinB)
=2sin(A+B2cos(A+B2)*2sin(A-B2)cos(A-B2)
=sin(A+B)sin(A-B)
Swarna Kamal Dhyawala ya this is more easier than my methodUpvote·0· Reply ·2013-02-05 08:30:45
3 Answers
996sin(A+B)sin(A--B)
= (sinAcosB+cosAsinB)(sinAcosB--cosAsinB)
= (sinAcosB)^2 -- (cosAsinB)^2
= sin^2A(1--sin^2B) --sin^2B(1--sin^2A)
= sin^2A -- sin^2Asin^2B -- sin^2B + sin^2Asin^2B
= sin^2A -- sin^2B
2305
36Also,
sin(A+B).sin(A-B)
= 1/2[cos2B - cos2A]
=1/2[1 - 2sin2B - 1 + 2sin2A]
= sin2A - sin2B.