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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
\(\frac{1-2sin^2x}{1-tanx}=\frac{cosx\left(1-2sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cos^2x-sin^2x\right)}{cosx-sinx}=\frac{cosx\left(cosx+sinx\right)\left(cosx-sinx\right)}{cosx-sinx}\)
\(=cosx\left(cosx+sinx\right)=\frac{cosx\left(cosx+sinx\right)^2}{cosx+sinx}=\frac{cos^2x+sin^2x+2sinx.cosx}{1+\frac{sinx}{cosx}}=\frac{1+sin2x}{1+tanx}\)
\(\frac{x}{2}=a\Rightarrow\frac{cot^2a-cot^23a}{cos^2a.cos2a\left(1+cot^23a\right)}=\frac{sin^23a\left(cot^2a-cot^23a\right)}{cos^2a.cos2a}=\frac{sin^23a.cot^2a-cos^23a}{cos^2a.cos2a}\)
\(=\frac{sin^23a.cos^2a-cos^23a.sin^2a}{sin^2a.cos^2a.cos2a}=\frac{\left(sin3a.cosa-cos3a.sina\right)\left(sin3a.cosa+cos3a.sina\right)}{sin^2a.cos^2a.cos2a}\)
\(=\frac{sin\left(3a-a\right).sin\left(3a+a\right)}{sin^2a.cos^2a.cos2a}=\frac{sin2a.sin4a}{sin^2a.cos^2a.cos2a}=\frac{2sina.cosa.4sina.cosa.cos2a}{sin^2a.cos^2a.cos2a}\)
\(=\frac{8sin^2a.cos^2a.cos2a}{sin^2a.cos^2a.cos2a}=8\)
\(sin\left(a+b+a\right)=5sin\left(a+b-a\right)\)
\(\Leftrightarrow sin\left(a+b\right)cosa+cos\left(a+b\right).sina=5sin\left(a+b\right).cosa-5cos\left(a+b\right).sina\)
\(\Leftrightarrow6cos\left(a+b\right).sina=4sin\left(a+b\right).cosa\)
\(\Leftrightarrow\frac{2sin\left(a+b\right)cosa}{cos\left(a+b\right)sina}=3\Leftrightarrow\frac{2tan\left(a+b\right)}{tana}=3\)
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
Lời giải:
a)
\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)
Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$
$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$
$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$
$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$
-------------------------
$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$
$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$
$\Rightarrow \cos 2a=\frac{-7}{25}$
$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$
$\Rightarrow \cot 2a=\frac{-7}{24}$
\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)
\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)
\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)
\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)
\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)
\(=sin2x.cosx-cos2x.sinx=sinx\)
\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)
\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)
\(sinx\left(1+cos2x\right)=sinx\left(1+2cos^2x-1\right)=2sinx.cosx.cosx=sin2x.cosx\)
\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=\frac{-cos2x}{\frac{1}{2}sin2x}=-\frac{2}{tan2x}\)
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}\left(\frac{1+cosx}{cosx}\right)=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}.\frac{2cos^2\frac{x}{2}}{cosx}=\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{cosx}=\frac{sinx}{cosx}=tanx\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\Rightarrow\) 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