A) 1-sin^x B) (1-cosx)(1+cosx) C) sinx-sinx.cos^x D) sin4x+cos4x+2sin^x.cos^x+tan^x.cos^x 17/07/2021 Bởi Sadie A) 1-sin^x B) (1-cosx)(1+cosx) C) sinx-sinx.cos^x D) sin4x+cos4x+2sin^x.cos^x+tan^x.cos^x
A. 1- $sin^{2}$x = $cos^{2}$x B. (1 – cosx).(1+ cosx) = 1- $cos^{2}$x = $sin^{2}$x C.sinx – sinx.$cos^{2}$x = sinx.( 1- $cos^{2}$x ) = $sin^{3}$x D. $sin^{4}$x + $cos^{4}$x + 2$sin^{2}$x.$cos^{2}$x + $tan{^2}$x.$cos^{2}$x = $(sin^{2}x)^{2}$ + $(cos^{2}x)^{2}$ + 2$sin^{2}$x.$cos^{2}$x $\frac{sin^{2}x}{cos^{2}x}$.$cos^{2}$x = 1 + $sin^{2}$x Bình luận
$a,1-\sin^2x=\cos^2x$ $b,(1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x$ $c,\sin x-\sin x.\cos^2x=\sin x.(1-\cos^2x)=\sin x.\sin^2x=\sin^3x$ $\begin{array}{l}d,\sin^4x+\cos^4x+2\sin^2x\cos^2x+\tan^2x.\cos^2x\\=(\sin^4x+2\sin^2x\cos^2x+\cos^4x)+\dfrac{\sin^2x}{\cos^2x}.\cos^2x\\=(\sin^2x+\cos^2x)^2+\sin^2x\\=1^2+\sin^2x\\=\sin^2x+1.\end{array}$ Bình luận
A. 1- $sin^{2}$x = $cos^{2}$x
B. (1 – cosx).(1+ cosx) = 1- $cos^{2}$x = $sin^{2}$x
C.sinx – sinx.$cos^{2}$x = sinx.( 1- $cos^{2}$x ) = $sin^{3}$x
D. $sin^{4}$x + $cos^{4}$x + 2$sin^{2}$x.$cos^{2}$x + $tan{^2}$x.$cos^{2}$x
= $(sin^{2}x)^{2}$ + $(cos^{2}x)^{2}$ + 2$sin^{2}$x.$cos^{2}$x $\frac{sin^{2}x}{cos^{2}x}$.$cos^{2}$x
= 1 + $sin^{2}$x
$a,1-\sin^2x=\cos^2x$
$b,(1-\cos x)(1+\cos x)=1-\cos^2x=\sin^2x$
$c,\sin x-\sin x.\cos^2x=\sin x.(1-\cos^2x)=\sin x.\sin^2x=\sin^3x$
$\begin{array}{l}d,\sin^4x+\cos^4x+2\sin^2x\cos^2x+\tan^2x.\cos^2x\\=(\sin^4x+2\sin^2x\cos^2x+\cos^4x)+\dfrac{\sin^2x}{\cos^2x}.\cos^2x\\=(\sin^2x+\cos^2x)^2+\sin^2x\\=1^2+\sin^2x\\=\sin^2x+1.\end{array}$