trig - prove identities

**getem j** · November 4th 2008, 11:47 AM

prove so L.S. = R.S.
sin8A = 8sinAcosAcos2Acos4A

please help have test tommorow

thanks

**Soroban** · November 4th 2008, 11:59 AM

Hello, getem j!

We're expected to know this identity: . $\sin2\theta \:=\:2\cdot\sin\theta\cdot\cos\theta$

Prove: . $\sin8A \:= \:8\cdot\sin A\cdot\cos A\cdot\cos2A\cdot\cos4A$

We have: . $\sin8A \;=\;2\cdot\underbrace{\sin4A}\cdot\cos4A$

. . . . . . . . . $=\; 2\overbrace{(2\cdot\sin2A\cdot\cos2A)}\cdot\cos4A$

. . . . . . . . . $=\;4\cdot\underbrace{\sin2A}\cdot\cos2A\cdot\cos4A$

. . . . . . $= \;4\cdot\overbrace{(2\cdot\sin A\cdot\cos A)}\cdot\cos2A\cdot\cos4A$

. . . . . . $= \;8\cdot\sin A\cdot\cos A\cdot\cos2A\cdot\cos4A$

**getem j** · November 4th 2008, 12:11 PM

wow thank you so much , your a life saver

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