[SOLVED] Verify that the equation is an identity.

**something3k** · October 7th 2008, 05:30 PM

$Sin 4\alpha = 4 Sin\alpha Cos\alpha Cos 2\alpha$

**Jhevon** · October 7th 2008, 05:32 PM

Originally Posted by something3k

$Sin 4\alpha = 4 Sin\alpha Cos\alpha Cos 2\alpha$

start on the left hand side and use the double angle formula for sine: $\sin 2 \alpha = 2 \sin \alpha \cos \alpha$

note that $\sin4 \alpha = \sin [2(2 \alpha)]$

**something3k** · October 7th 2008, 05:48 PM

so would that equal $4 \sin\alpha \cos\alpha?$

**Jhevon** · October 7th 2008, 06:22 PM

Originally Posted by something3k

so would that equal $4 \sin\alpha \cos\alpha?$

no, follow the rule: sin(2A) = 2*sinA*cosA

**Soroban** · October 7th 2008, 07:02 PM

Hello, something3k!

Identity: . $2\!\cdot\!\sin\theta\!\cdot\!\cos\theta \:=\:\sin2\theta$

$\sin4\alpha \:= \:4\!\cdot\!\sin\alpha\!\cdot\!\cos\alpha\!\cdot\! \cos2\alpha$

Or start on the right side . . .

$4\!\cdot\!\sin\alpha\!\cdot\!\cos\alpha\!\cdot\!\c os2\alpha \;=\;2\cdot\underbrace{2\!\cdot\!\sin\alpha\!\cdot \!\cos\alpha}_{\text{This is }\sin2\alpha}\cdot\cos2\alpha$

. . . . . . . . . . . . . $= \;\underbrace{2\!\cdot\!\sin2\alpha\!\cdot\!\cos2\ alpha}$

. . . . . . . . . . . . . $=\qquad \sin4\alpha$

**something3k** · October 7th 2008, 07:11 PM

ahh i seee, thanks a lot you guys i am thinking a little too hard.

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