1. ## Simplifying the expression

hey guys having trouble simplying this expression. Any help? thank u

$\displaystyle 2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x$

2. Use the double angle formulae for sin and cosine.

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

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

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

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

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

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

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

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

Do you see the pattern here?

4. Originally Posted by Chop Suey

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

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

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

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

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

Do you see the pattern here?
yes yes i see it. so using these double angled formulas i answer this question

5. Originally Posted by Chop Suey

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

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

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

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

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

Do you see the pattern here?
ok so:
$\displaystyle 2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x$

simplifys to:

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

this correct?

6. 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:

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

7. 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:

$\displaystyle \sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}$
yes yes thats right.
so its

sin6xcos5x-cos6xsin5x

?

8. Originally Posted by jvignacio
yes yes thats right.
so its

sin6xcos5x-cos6xsin5x

?
Yes. Don't doubt yourself.

And don't forget:
$\displaystyle \sin{(A\pm B)} = \sin{A}\cos{B} \pm \cos{A}\sin{B}$

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

And don't forget:
thanks man, will do. appreciate the help