CMR: `cos^4x-sin^4x=2cos^2x-1=1-2sin^2x` 29/08/2021 Bởi Natalia CMR: `cos^4x-sin^4x=2cos^2x-1=1-2sin^2x`
$\cos^4x-\sin^4x= (\cos^2x-\sin^2x)(\cos^2x+\sin^2x)= \cos^2x-\sin^2x$ $+) \cos^2x-\sin^2x= \cos^2x-(1-\cos^2x)= 2\cos^2x-1$ $+) \cos^2x-\sin^2x= 1-\sin^2x-\sin^2x=1-2\sin^2x$ Vậy $\cos^4x-\sin^4x= 2\cos^2x-1=1-2\sin^2x$ Bình luận
Ta có: VT= ( cos² x – sin²x )( cos²x + sin²x ) mà sin²x + cos²x = 1 với mọi x ⇒VT= cos² x – sin²x = 2cos² x – ( cos²x + sin²x) = 2cos² x -1 ⇒ cos² x – sin²x = cos²x + sin²x – 2sin²x = 1 – 2sin²x (đpcm) *Chúc bạn học tốt Bình luận
$\cos^4x-\sin^4x= (\cos^2x-\sin^2x)(\cos^2x+\sin^2x)= \cos^2x-\sin^2x$
$+) \cos^2x-\sin^2x= \cos^2x-(1-\cos^2x)= 2\cos^2x-1$
$+) \cos^2x-\sin^2x= 1-\sin^2x-\sin^2x=1-2\sin^2x$
Vậy $\cos^4x-\sin^4x= 2\cos^2x-1=1-2\sin^2x$
Ta có:
VT= ( cos² x – sin²x )( cos²x + sin²x ) mà sin²x + cos²x = 1 với mọi x
⇒VT= cos² x – sin²x = 2cos² x – ( cos²x + sin²x) = 2cos² x -1
⇒ cos² x – sin²x = cos²x + sin²x – 2sin²x
= 1 – 2sin²x (đpcm)
*Chúc bạn học tốt