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\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt
không phụ thuộc vào x.
a) \(sin^4x+cos^4x=\left(sin^2x\right)^2+\left(cos^2x\right)^2\)
\(=\left(sin^2x\right)^2+2sin^2xcos^2x+\left(cos^2x\right)^2-2sin^2xcos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\)
b) \(\dfrac{1+cotx}{1-cotx}=\dfrac{tanx.cotx+cotx}{tanx.cotx-cotx}\)
\(=\dfrac{cotx.\left(tanx+1\right)}{cotx.\left(tanx-1\right)}\)
\(=\dfrac{tanx+1}{tanx-1}\)
c) \(\dfrac{cosx+sinx}{cos^3x}=\dfrac{1}{cos^2x}+\dfrac{tanx}{cos^2x}\)
\(=1+tan^2x+tanx.\dfrac{1}{cos^2x}\)
\(=1+tan^2x+tanx.\left(1+tan^2x\right)\)
\(=1+tan^2x+tanx+tan^3x\)
\(=tan^3x+tan^2x+tanx+1\)
Lời giải:
a.
$\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-2\sin ^2x\cos ^2x$
b.
$\frac{1+\cot x}{1-\cot x}=\frac{1+\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}=\frac{\cos x+\sin x}{\sin x-\cos x}(1)$
$\frac{\tan x+1}{\tan x-1}=\frac{\frac{\sin x}{\cos x}+1}{\frac{\sin x}{\cos x}-1}=\frac{\cos x+\sin x}{\sin x-\cos x}(2)$
Từ $(1); (2)$ ta có đpcm
c.
$\frac{\cos x+\sin x}{\cos ^3x}=(1+\frac{\sin x}{\cos x}).\frac{1}{\cos ^2x}$
$=(1+\tan x).\frac{\sin ^2x+\cos ^2x}{\cos ^2x}$
$=(1+\tan x)(\tan ^2x+1)=\tan ^3x+\tan ^2x+\tan x+1$
Ta có đpcm.
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)
\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)
\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)
\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)
\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)
\(=sin2x.cosx-cos2x.sinx=sinx\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\Rightarrow\) đề sai
b/
\(\frac{1-cos4x}{sin4x}=\frac{1-\left(1-2sin^22x\right)}{2sin2x.cos2x}=\frac{2sin^22x}{2sin2x.cos2x}=\frac{sin2x}{cos2x}=tan2x\)
Đề sai tiếp lần 2
xin lỗi, mk nhấn thiếu chỗ tan2x