1-2sin^2x+sin^4x
I believe the answer is cos^4x
I want to say that 1-2(1+sin^2x)(sin^2x) if I move it around I get
I do not know where to go next or how it is cos^4x
As it says "factor" in your thread title...
$\displaystyle y=sin^2x$
and using $\displaystyle sin^2x+cos^2x=1\Rightarrow\ sin^2x=1-cos^2x$
$\displaystyle 1-2y+y^2=(y-1)(y-1)=\left(sin^2x-1\right)\left(sin^2x-1\right)=\left(1-cos^2x-1\right)\left(1-cos^2x-1\right)$
which leads to the answer.
Hello, IDontunderstand!
$\displaystyle \text{Sinplifyt: }\:1-2\sin^2\!x+\sin^4\!x$
$\displaystyle \text{I believe the answer is: }\:\cos^4\!x$
Factor: .$\displaystyle 1 - 2\sin^2\!x + \sin^4\!x \;=\;(\underbrace{1 - \sin^2\!x}_{\text{This is }\cos^2\!x})^2 \;=\;(\cos^2\!x)^2 \;=\;\cos^4\!x$