Factoring Trigonometric Expression

Hello Everyone,

I have to factor the following expression and I can't figure out the proper procedures to do it (Headbang).

I'm assuming that I must take the difference of squares?

$\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)$

Re: Factoring Trigonometric Expression

Remove the 1/3 as a common factor and then make use of $\displaystyle (a-1)^{2}\equiv a^{2} - 2a + 1$ followed by the trig identity $\displaystyle \sin ^{2}A + \cos^{2}A=1.$

Re: Factoring Trigonometric Expression

Quote:

Originally Posted by

**vaironxxrd** I have to factor the following expression and I can't figure out the proper procedures to do it .

$\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)$

$\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)=\frac{(1-sin^2(x))^2}{3}$.

Re: Factoring Trigonometric Expression

Quote:

Originally Posted by

**Plato** $\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)=\frac{(1-sin^2(x))^2}{3}$.

Plato is the following acceptable?

$\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x) = \frac{1}{3}(1-2sin^2(x)+sin^4(x)) $

Re: Factoring Trigonometric Expression

Quote:

Originally Posted by

**vaironxxrd** Plato is the following acceptable?

$\displaystyle \frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x) = \frac{1}{3}(1-2sin^2(x)+sin^4(x)) $

**Acceptable to whom?**

How would I know that answer to that?