Proving identities

**arze** · July 9th 2009, 06:44 PM

Prove (sin2A-sinA)(1+2cosA)=sin3A
i used the identity sinA-sinB=2cos(A+B)/2*sin(A-B)/2, but then i got stuck.
Thanks

**Soroban** · July 9th 2009, 07:50 PM

Hello, arze!

This requires Olympic-level gymnastics . . .

Prove: . $(\sin2A-\sin A)(1+2\cos A) \:=\:\sin3A$

$(\sin2x - \sin A)(1 + 2\cos A) \;=\;(2\sin A\cos A - \sin A)(1 + 2\cos A)$

. . . . . $=\;2\sin A\cos A + 4\sin A\cos^2\!A - \sin A - 2\sin A\cos A$

. . . . . $- \;4\sin A\underbrace{\cos^2\!A}_{\searrow} - \sin A$
. . . . . $= \;4\sin A(1 - \sin^2\!A) - \sin A$

. . . . . $= \;4\sin A - 4\sin^3\!A - \sin A$

. . . . . $= \qquad\underbrace{3\sin A} \qquad- \qquad \underbrace{4\sin^3\!A}$
. . . . . $= \;\overbrace{2\sin A + \sin A} - \overbrace{2\sin^3\!A - 2\sin^3\!A}$

. . . . . $= \;2\sin A-2\sin^3\!A + \sin A - 2\sin^3\!A$

. . . . . $= \;2\sin A\underbrace{(1 - \sin^2\!A)} \;+ \;\sin A\underbrace{(1 - 2\sin^2\!A)}$
. . . . . $= \quad2\sin A\overbrace{\cos^2\!A} \qquad+ \qquad \sin A\overbrace{\cos2A}$

. . . . . $= \;\underbrace{(2\sin A\cos A)}\cos A + \sin A\cos2A$
. . . . . $=\qquad \overbrace{\sin2A}\cos A + \sin A\cos2A$

. . . . . $= \qquad\qquad\qquad\sin3A$

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