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không phụ thuộc vào x.
\(\frac{sin2x-sin4x}{1-cos2x+cos4x}=\frac{sin2x-2sin2x.cos2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(1-2cos2x\right)}{-cos2x\left(1-2cos2x\right)}=\frac{-sin2x}{cos2x}=-tan2x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=-\left(\frac{sin2x-sin4x}{1-cos2x+cos4x}\right)=-\left(-tan2x\right)=tan2x\) lấy luôn kết quả câu trên cho lẹ, biến đổi thì làm y hệt
\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)
\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)
Chọn A.
Từ giả thiết suy ra:
A = (cos4x + cos2x sin2x) + sin2x = cos2x(sin2x + cos2x ) + sin2x
A = cos2x.1 + sin2x = 1
Rút gọn
A= \(\frac{cosx-cos2x-cos3x+cos4x}{sinx-sin2x-sin3x+sin4x}\)
B= sinx(1+2cos2x+2cos4x+2cos6x)
\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)
\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)
\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)
\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)
\(=sin7x\)
\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)
\(\Rightarrow\) 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
\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)
b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)
c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)
d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)
\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)
e/
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