# Factoring

• Apr 18th 2009, 07:59 PM
vrod guy
Factoring
I am supposed to factor a trig expression.

the expression is:

1-2 sin^2x+sin^4x

Any help is appreciated!
• Apr 18th 2009, 08:04 PM
Jhevon
Quote:

Originally Posted by vrod guy
I am supposed to factor a trig expression.

the expression is:

1-2 sin^2x+sin^4x

Any help is appreciated!

note that your expression is a quadratic in $\displaystyle \sin^2 x$. to see this, let $\displaystyle a = \sin^2 x$, then your expression is

$\displaystyle 1 - 2a + a^2$

now hopefully you can factor that, then just replace $\displaystyle a$ with $\displaystyle \sin^2 x$ when done
• Apr 19th 2009, 06:36 AM
scottsw1
Jhevon,
When I factored, I came up with (1-sin^2x)^2=cos^4x. Is this correct? Vrod guy, I hope I was able to come up with the correct answer for you.
• Apr 20th 2009, 06:09 PM
vrod guy
Scott, Are you sure this is correct? Does anyone else agree with Scott?
• Apr 20th 2009, 06:24 PM
Prove It
Not if you were asked to factorise.

$\displaystyle 1 - 2a + a^2 = (1 - a)^2$

$\displaystyle = (1 - \sin^2{x})^2$

$\displaystyle = \left[(1 + \sin{x})(1 - \sin{x})\right]^2$.