Trig

**meli3000** · June 12th 2008, 02:24 PM

prove the identity

sin 3 x = 3 sin x - 4 sin ^3 x

**TheEmptySet** · June 12th 2008, 03:40 PM

Originally Posted by meli3000

prove the identity

sin 3 x = 3 sin x - 4 sin ^3 x

See this thread

http://www.mathhelpforum.com/math-he...ease-help.html

**Soroban** · June 12th 2008, 03:53 PM

Hello, meli3000!

Prove the identity: . $\sin3x \:= \:3\sin x - 4\sin^3\!x$

We will need these identites:
. . $\sin(A+B) \;=\;\sin A\cos B + \sin B\cos A$
. . $\sin2A \:=\: 2\sin A\cos A$
. . $\cos2A \:=\:1-2\sin^2\!A$
. . $\cos^2\!A \:=\:1 - \sin^2\!A$

$\sin(3x) \;=\;\sin(2x + x)$

. . . . . $= \;\sin(2x)\cos(x) + \sin(x)\cos(2x)$

. . . . . $= \;(2\sin x\cos x)\cos x + \sin x(1 - 2\sin^2\!x)$

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

. . . . . $= \;2\sin x(1 - \sin^2\!x) + \sin x - 2\sin^3\!x$

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

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

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