Question

How do you factor and simplify `sin^4x-cos^4x`?

Answer 1

`(sinx-cosx)(sinx+cosx)`

Explanation:

Factorizing this algebraic expression is based on this property:

`a^2 - b^2 =(a - b)(a + b)`

Taking `sin^2x =a` and `cos^2x=b` we have :

`sin^4x-cos^4x=(sin^2x)^2-(cos^2x)^2=a^2-b^2`

Applying the above property we have:

`(sin^2x)^2-(cos^2x)^2=(sin^2x-cos^2x)(sin^2x+cos^2x)`

Applying the same property on`sin^2x-cos^2x`

thus,

`(sin^2x)^2-(cos^2x)^2`
`=(sinx-Cosx)(sinx+cosx)(sin^2x+cos^2x)`

Knowing the Pythagorean identity, `sin^2x+cos^2x=1` we simplify the expression so,

`(sin^2x)^2-(cos^2x)^2`
`=(sinx-Cosx)(sinx+cosx)(sin^2x+cos^2x)`
`=(sinx-cosx)(sinx+cosx)(1)`
`=(sinx-cosx)(sinx+cosx)`

Therefore,
`sin^4x-cos^4x=(sinx-cosx)(sinx+cosx)`

Answer 2

= - cos 2x

Explanation:

`sin^4x - cos^4 x = (sin^2 x + cos ^2 x)(sin^2 x - cos^2 x)`
Reminder:
`sin^2 x + cos^2 x = 1`, and
`cos^2 x - sin^2 x = cos 2x`
Therefore:
`sin^4x - cos^4 x = - cos 2x`