simplifying identities

**shawli** · January 4th 2010, 04:12 PM

Simplify:

$\frac{sin6x}{sin2x} - \frac{cos6x}{cos2x}$

How do i simplify $sin6x$ and $cos6x$ ?

**Krizalid** · January 4th 2010, 04:14 PM

Hint: $\sin (4x)=\sin (6x-2x)=\sin (6x)\cos (2x)-\cos (6x)\sin (2x),$ and $\sin(4x)=2\sin(2x)\cos(2x).$

**Soroban** · January 4th 2010, 05:21 PM

Hello, shawli!

Simplify: . $\frac{\sin6x}{\sin2x} - \frac{\cos6x}{\cos2x}$

$\text{Combine the fractions: }\;\frac{\overbrace{\sin6x\cos2x - \cos6x\sin2x}^{\text{This is }\sin(6x-2x)}}{\sin2x\cos2x} \;=\;\frac{\sin4x}{\sin2x\cos2x}$

. . . . . . . . . . . . . $=\frac{\sin4x}{\frac{1}{2}(\underbrace{2\sin2x\cos 2x)}_{\text{This is }\sin4x}} \;=\;\frac{\sin4x}{\frac{1}{2}\sin4x} \;=\;2$

**Krizalid** · January 4th 2010, 05:23 PM

You screwed out my Hint.

Next time let them to proceed with the Hint.

**shawli** · January 4th 2010, 05:23 PM

thank you both!

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