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\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\Rightarrow\) 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
\(A=\frac{1}{2}\left(\frac{sin^2x}{cos^2x}-1\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)
\(A=\frac{-1}{2}\left(\frac{cos^2x-sin^2x}{cos^2x}\right)\frac{cosx}{sinx}+cos4x.cot2x+sin4x\)
\(A=\frac{-cos2x}{2cosx.sinx}+cos4x.cot2x+sin4x\)
\(A=-cot2x+cos4x.cot2x+sin4x\)
\(A=cot2x\left(cos4x-1\right)+sin4x\)
\(A=\frac{cos2x}{sin2x}.\left(1-2sin^22x-1\right)+sin4x\)
\(A=\frac{-2cos2x.sin^22x}{sin2x}+sin4x\)
\(A=-sin4x+sin4x=0\)
\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
không phụ thuộc vào x.
\(\frac{1}{sin2a}=\frac{sina}{sina.sin2a}=\frac{sin\left(2a-a\right)}{sina.sin2a}=\frac{sin2a.cosa-cos2a.sina}{sina.sin2a}\)
\(=\frac{sin2a.cosa}{sina.sin2a}-\frac{cos2a.sina}{sina.cos2a}=\frac{cosa}{sina}-\frac{cos2a}{sin2a}=cota-cot2a\)
Áp dụng vào bài toán:
\(A=\frac{1}{sin2y}+\frac{1}{sin2\left(2y\right)}+\frac{1}{sin2\left(4y\right)}-coty+cot8y\)
\(=coty-cot2y+cot2y-cot4y+cot4y-cot8y-coty+cot8y\)
\(=0\)
\(B=\frac{1}{sin2\left(2x\right)}+\frac{1}{sin2\left(2x\right)}+\frac{1}{sin2\left(8x\right)}-cot2x+cot16x\)
\(=cot2x-cot4x+cot4x-cot8x+cot8x-cot16x-cot2x+cot16x\)
\(=0\)