1. ## identity showing practice

hi guys, im tackling this question and need abit of guidence.

$
sin2x + sin4x + sin6x = 4cosxcos2xsin3x
$

so far i got:

( $2sinxcosx)+(4sin2xcos2x)+(6sin3xcos3x)$

not sure where to go after this?

any help much appreciated

2. Continue simplifying the double angles (I can see two in the middle part that can be done right now), and see if you can add the three together.

3. Originally Posted by atac777
Continue simplifying the double angles (I can see two in the middle part that can be done right now), and see if you can add the three together.

okay umm

$2sinxcosx + 2(2sinxcosx) + 3(2sinxcosx)$

then i dont know what to do.. i get confused easy

4. Originally Posted by jvignacio
hi guys, im tackling this question and need abit of guidence.

$
sin2x + sin4x + sin6x = 4cosxcos2xsin3x
$

so far i got:

( $2sinxcosx)+(4sin2xcos2x)+(6sin3xcos3x)$

not sure where to go after this?

any help much appreciated
$\sin 4x = 2 \sin2x \cos2x$ and $\sin 6x = 2 \sin3x \cos3x$

Now

$\sin 6x = 2 \sin 3x \cos 3x = 2 \sin 3x (\cos 2x \cos x - \sin 2x \sin x)$

$\sin 6x = 2 \cos 2x \cos x \sin 3x - 2 \sin 2x \sin x \sin 3x$

The first part is OK (same form as the second member to find)

The second part is $- 2 \sin 2x \sin x \sin 3x = -2 \sin 2x \sin x (\sin 2x \cos x + \sin x \cos 2x)$

$- 2 \sin 2x \sin x \sin 3x = -2 \sin^2 2x \sin x \cos x - 2 \sin^2 x \cos 2x \sin 2x$

$- 2 \sin 2x \sin x \sin 3x = -2 (1 - \cos^2 2x) \sin x \cos x - 2 (1 - \cos^2 x) \cos 2x \sin 2x$

Expanding and using $2 \sin x \cos x = \sin 2x$ and $2 \sin2x \cos2x = \sin 4x$

$- 2 \sin 2x \sin x \sin 3x = - \sin 2x + 2 \cos^2 2x \sin x \cos x - \sin 4x + 2 \cos^2 x \cos 2x \sin 2x$

$- 2 \sin 2x \sin x \sin 3x = - \sin 2x - \sin 4x + 2 \cos 2x \cos x (\cos 2x \sin x + \cos x \sin 2x)$

$- 2 \sin 2x \sin x \sin 3x = - \sin 2x - \sin 4x + 2 \cos 2x \cos x \sin 3x$

Therefore

$\sin 6x = 2 \cos 2x \cos x \sin 3x - \sin 2x - \sin 4x + 2 \cos 2x \cos x \sin 3x$

$\sin 2x + \sin 4x + \sin 6x = 4 \cos x \cos 2x \sin 3x$