Question

`sin^4x-cos^4x=1-2cos^2x` prove it?

Answer 1

We want to show that `sin^4x-cos^4x=1-2cos^2x`

We'll work with the LHS:
Using the identity `sin^2x+cos^2x-=1` we get:
`(1-cos^2x)^2-cos^4x`

`1-2cos^2x+cos^4x-cos^4x`

`1-2cos^2x`

`LHS=1-2cos^2x`

`LHS=RHS`

Answer 2

See explanation...

Explanation:

We will use Pythagoras' identity:

`sin^2 x + cos^2 x = 1`

from which we can deduce:

`sin^2 x = 1 - cos^2 x`

Also note that the difference of squares identity can be written:

`A^2-B^2 = (A-B)`

We can use this with `A=sin^2 x` and `B=cos^2 x` as follows:

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

`color(white)(sin^4 x - cos^4 x) = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x)`

`color(white)(sin^4 x - cos^4 x) = sin^2 x - cos^2 x`

`color(white)(sin^4 x - cos^4 x) = (1-cos^2 x) - cos^2 x`

`color(white)(sin^4 x - cos^4 x) = 1-2cos^2 x`