Factoring Trigonometric Expression

**vaironxxrd** · February 18th 2013, 06:17 AM

Hello Everyone,
I have to factor the following expression and I can't figure out the proper procedures to do it

.

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

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

**BobP** · February 18th 2013, 07:44 AM

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

**Plato** · February 18th 2013, 08:15 AM

Originally Posted by vaironxxrd

I have to factor the following expression and I can't figure out the proper procedures to do it .
$\frac{1}{3} - \frac{2}{3}sin^2(x) + \frac{1}{3}sin^4(x)$

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

**vaironxxrd** · February 18th 2013, 11:01 AM

Originally Posted by Plato

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

Plato is the following acceptable?

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

**Plato** · February 18th 2013, 11:08 AM

Originally Posted by vaironxxrd

Plato is the following acceptable?

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

Acceptable to whom?

How would I know that answer to that?

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