prove that

2sin2a - 3cos2a - 3sina + 3 = sina(4cosa + 6sina - 3)

i am very tired and cant seem to figure it out. would be grateful for any help.

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- Mar 16th 2008, 09:45 AMOranges&Lemonsproving an identity
prove that

2sin2a - 3cos2a - 3sina + 3 = sina(4cosa + 6sina - 3)

i am very tired and cant seem to figure it out. would be grateful for any help. - Mar 16th 2008, 10:03 AMMoo
Hello,

sin(2a)=2cos(a)sin(a)

cos(2a)=cosē(a)-sinē(a)=(1-sinē(a))-sinē(a)=1-2sinē(a)

If you replace these values in the initial expression, you'll get what you want ;-)