how would you prove sin4x - cos4x = sin2x-cos2x as well as sin4x + 2sin2xcos2x +cos4x = 1
:confused:
![]() |
how would you prove sin4x - cos4x = sin2x-cos2x as well as sin4x + 2sin2xcos2x +cos4x = 1
:confused:
In both cases, you'll want to factor the left side of the equation. Treat sin^4 x - cos^4 x as a difference of two squares and treat sin^4 x + 2sin^2x cos^2x + cos^4x as a perfect-square trinomial. Also the 1st. Pythagorean Identity will come into play ; sin^2x + cos^2x = 1.
I think you have written the question incorrectly.
These identities are not correct.
I think you should have written:
1. (Sinx)^4 - (Cos x)^4
= [ (sinx)^2 - (cos)^2 ] [ (sinx)^2 + (Cos x)^2) ]
= (sinx)^2 - (cos)^2
2. sin^4 x + 2 sin^2 co^2 x + cos^4 x
= (sin^2 x + cos^2 x)^2
= 1^2
= 1
All times are GMT -7. The time now is 07:21 PM. |