# Simplifying the expression

• Sep 9th 2008, 07:33 AM
jvignacio
Simplifying the expression
hey guys having trouble simplying this expression. Any help? thank u

$

2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x

$
• Sep 9th 2008, 08:03 AM
Krizalid
Use the double angle formulae for sin and cosine.

(Sorry for my poor help but I'm in a rush.)
• Sep 9th 2008, 08:08 AM
Chop Suey
Quote:

Originally Posted by Krizalid
Use the double angle formulae for sin and cosine.

(Sorry for my poor help but I'm in a rush.)

$\cos{2x} = \cos^2{x} - \sin^2{x}$

$\cos{4x} = \cos^2{2x} - \sin^2{2x}$
$\ldots$

==================

$\sin{2x} = 2\sin{x}\cos{x}$

$\sin{4x} = 2\sin{2x}\cos{2x}$
$\ldots$

Do you see the pattern here?
• Sep 9th 2008, 08:19 AM
jvignacio
Quote:

Originally Posted by Chop Suey

$\cos{2x} = \cos^2{x} - \sin^2{x}$

$\cos{4x} = \cos^2{2x} - \sin^2{2x}$
$\ldots$

==================

$\sin{2x} = 2\sin{x}\cos{x}$

$\sin{4x} = 2\sin{2x}\cos{2x}$
$\ldots$

Do you see the pattern here?

yes yes i see it. so using these double angled formulas i answer this question
• Sep 9th 2008, 08:34 AM
jvignacio
Quote:

Originally Posted by Chop Suey

$\cos{2x} = \cos^2{x} - \sin^2{x}$

$\cos{4x} = \cos^2{2x} - \sin^2{2x}$
$\ldots$

==================

$\sin{2x} = 2\sin{x}\cos{x}$

$\sin{4x} = 2\sin{2x}\cos{2x}$
$\ldots$

Do you see the pattern here?

ok so:
$2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x$

simplifys to:

$
sin6xcos^2\frac{5}{2}x-sin^2\frac{5}{2}x-(cos6x)(2sin\frac{5}{2}xcos\frac{5}{2}x)
$

this correct?
• Sep 9th 2008, 08:40 AM
Chop Suey
You only needed to change the 2sin3xcos3x and (cos^2(3x) - sin^2(3x)).

Then, you'll see that this fits the sum identitiy for sine. Recall that:

$\sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}$
• Sep 9th 2008, 08:43 AM
jvignacio
Quote:

Originally Posted by Chop Suey
You only needed to change the 2sin3xcos3x and (cos^2(3x) - sin^2(3x)).

Then, you'll see that this fits the sum identitiy for sine. Recall that:

$\sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}$

yes yes thats right.
so its

sin6xcos5x-cos6xsin5x

?
• Sep 9th 2008, 08:45 AM
Chop Suey
Quote:

Originally Posted by jvignacio
yes yes thats right.
so its

sin6xcos5x-cos6xsin5x

?

Yes. Don't doubt yourself.

And don't forget:
Quote:

$\sin{(A\pm B)} = \sin{A}\cos{B} \pm \cos{A}\sin{B}$
• Sep 9th 2008, 08:58 AM
jvignacio
Quote:

Originally Posted by Chop Suey
Yes. Don't doubt yourself.

And don't forget:

thanks man, will do. appreciate the help