Sin ^4x-cos^4x=1-2cos^2x

Question

Need help proving this identity

Answer 1

Use a⁴ - b⁴ = (a² -b²)(a²+b²)  - difference of two squares-

and the identities

cos 2x =cos²x - sin²x = 2 cos ² x - 1

sin²x + cos²x = 1,

and you would be able to prove it.

Answer 2

multiply the whole problem by -1

cos^4x - sin^4x= 2cos^2x - 1

 

things to remember:

cos(2x) = 2cos^(X) - 1

 

sin(X/2)  = sqrt(1-cos(X) / 2) ....> then sin^2(x) =  1/2(1-cos(2x)

cos(X/2) = sqrt(1+cos(X) / 2) ....>   then cos^2(x) = 1/2(1-cos(2x)

Then....

 

On the left side:

cos^4x - sin^4x =  [1/2(1+cos(2x)]^2 - [1/2(1-cos(2x)]^2 and simplify 

= cos(2x) = 2cos^(X) - 1 = right side 

 

good luck!