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\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)
\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)
\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)
\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)
\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)
\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)
\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)
b/ \(\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a+1\right)}{2\left(cos^22a+2cos2a+1\right)}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}\)
\(\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{4sin^4a}{4cos^4a}=tan^4a\)
c/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)
\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)
\(=1-sin^2x=cos^2x\)
xét vế phải
( cosa+1-sina)^2
= cos^2 +1+ sin^2+2cosa-2sina-2sinacosa
= 2( 1+ cosa-sina-sinacosa)
= 2( 1-sina) ( 1+cosa)
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
\(sina\sqrt{1+\frac{sin^2a}{cos^2a}}=sina\sqrt{\frac{cos^2a+sin^2a}{cos^2a}}=\frac{sina}{\left|cosa\right|}=\pm tana\)
\(\frac{1-cos^2x}{1-sin^2x}+tanx.cotx=\frac{sin^2x}{cos^2x}+\frac{sinx}{cosx}.\frac{cosx}{sinx}=tan^2x+1=\frac{1}{cos^2x}\)
\(\frac{1-4sin^2xcos^2x}{\left(sinx+cosx\right)^2}=\frac{\left(1-2sinx.cosx\right)\left(1+2sinx.cosx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(1-sin2x\right)\left(1+2sinx.cosx\right)}{1+2sinx.cosx}=1-2sinx\)
\(sin\left(90-x\right)+cos\left(180-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)
\(=cosx-cosx+sin^2x.\frac{1}{cos^2x}-tan^2x=tan^2x-tan^2x=0\)
Giải các Pt sau:
cos5s - sin2x =0
sin5x + cos2x =1
cos2x + \(2\sqrt{3}sinxcosx\) - sin2x = \(\sqrt{2}\)
Ủa câu này nãy làm rồi mà bạn chưa hiểu hay sao?
\(VT=cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^{2x}\)
Ở đây ta lần lượt có:
\(cos^22x+sin^22x=1\)
\(2sin2x.cos2x=sin4x\)
\(2sin3x.cosx=sin4x+sin2x\)
\(2sinx.cosx=sin2x\)
Ghép lại sẽ được:
\(VT=1-sin4x+sin4x+sin2x-sin2x-sin^2x=1-sin^2x=cos^2x\)