chứng minh công thức lg -----

P

pe_lun_hp

Hế hế, mình phục mình ghê hớ hớ

$\dfrac{sin^4x + 2cosxsinx - cos^4x}{tan2x - 1} = cos2x$

Tử:

$sin^4x + 2cosxsinx - cos^4x = (sin^4x - cos^4x) + 2cosxsinx$

$ = (sin^2x - cos^2x) + 2cosxsinx$

$ = -(cos^2x - sin^2x) + 2cosxsinx$

$ = -2cosx^2 + 1 + 2cosxsinx$ (1)

Mẫu:

$tan2x - 1 = \dfrac{sin2x}{cos2x} - 1$

$ = \dfrac{2cosx.sinx}{2cos^2x - 1} - 1$

$= \dfrac{2cosxsinx - 2cos^2x + 1}{2cos^2x - 1}$ (2)

Từ (1),(2) ta có:

$\dfrac{sin^4x + 2cosxsinx - cos^4x}{tan2x - 1} = \dfrac{-2cosx^2 + 1 + 2cosxsinx}{ \dfrac{2cosxsinx - 2cos^2x + 1}{2cos^2x - 1}} = cos2x$ (đpcm)


keke là kaka =))
 
You must log in or register to reply here.
Top Bottom