Answer:
[tex]2\sin^2{x}-\sin^4{x}[/tex]
Step-by-step explanation:
I'll be utilizing two identities here: the difference of squares identity [tex]a^2-b^2=(a+b)(a-b)[/tex], and the trigonometric identity [tex]\sin^2{x}+\cos^2{x}=1[/tex].
To start, we can rewrite [tex]1-\cos^4{x}[/tex] as the difference of squares [tex]1^2-(\cos^2{x})^2[/tex], which lets us rewrite it as [tex](1+\cos^2{x})(1-\cos^2{x})[/tex].
We can subtract [tex]\sin^2{x}[/tex] from both sides of the identity [tex]\sin^2{x}+\cos^2{x}=1[/tex] to find that [tex]\cos^2{x}=1-\sin^2{x}[/tex], and similarly, subtracting [tex]\cos^2{x}[/tex] gives us the equation [tex]\sin^2{x} = 1-\cos^2{x}[/tex]. We can use these two facts to make some substitutions in our expression, so that
[tex](1+\cos^2{x})(1-\cos^2{x}) = (1+1-\sin^2{x})(\sin^2{x})=(2-\sin^2{x})(\sin^2{x})[/tex]
Simplifying this expression:
[tex](2-\sin^2{x})(\sin^2{x})\\=2\sin^2{x}-(\sin^2{x})^2\\=2\sin^2{x}-\sin^4{x}[/tex]
So our solution is [tex]2\sin^2{x}-\sin^4{x}[/tex]!
