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ơ bạn :\(\dfrac{cos\left(x+y\right)+cosx}{cos\left(x+y\right)-cosx}=\dfrac{2cos\left(\dfrac{2x+y}{2}\right).cos\left(\dfrac{y}{2}\right)}{-2sin\left(\dfrac{2x+y}{2}\right).sin\left(\dfrac{y}{2}\right)}=-2.cot\left(\dfrac{2x+y}{2}\right).cot\left(\dfrac{y}{2}\right)\) L không thể bẳng 0 được
Lời giải:
$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$
$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$
\(x+2y=\dfrac{\pi}{2}\)
\(\Leftrightarrow x+y=\dfrac{\pi}{2}-y\) thay vào A được:
\(A=\dfrac{cos\left(\dfrac{\pi}{2}-y\right)-cosy}{cos\left(\dfrac{\pi}{2}-y\right)+cosy}\)\(=\dfrac{siny-cosy}{siny+cosy}\)\(=\dfrac{\dfrac{\sqrt{2}}{2}.siny-\dfrac{\sqrt{2}}{2}.cosy}{\dfrac{\sqrt{2}}{2}.siny+\dfrac{\sqrt{2}}{2}cosy}\)\(=\dfrac{cos\dfrac{\pi}{4}.siny-sin\dfrac{\pi}{4}.cosy}{sin\dfrac{\pi}{4}.siny+cos\dfrac{\pi}{4}.cosy}\)
\(=\dfrac{sin\left(y-\dfrac{\pi}{4}\right)}{cos\left(y-\dfrac{\pi}{4}\right)}\)\(=tan\left(y-\dfrac{\pi}{4}\right)\)
\(x+2y=\dfrac{\pi}{2}\Rightarrow x+y=\dfrac{\pi}{2}-y\)
\(\Rightarrow cos\left(x+y\right)=cos\left(\dfrac{\pi}{2}-y\right)\)
\(\Rightarrow cos\left(x+y\right)=siny\)
Do đó: \(A=\dfrac{siny-cosy}{siny+cosy}=\dfrac{\sqrt{2}sin\left(y-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(y-\dfrac{\pi}{4}\right)}=tan\left(y-\dfrac{\pi}{4}\right)\)
\(P=\frac{sin^2x+cos^2x+2sinx.cosx-1}{\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right).cotx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sinx.cosx}{\left(cosx-sinx\right).\frac{cosx}{sinx}}-\frac{1}{cosx-sinx}=\frac{2sin^2x}{cosx-sinx}-\frac{1}{cosx-sinx}\)
\(=\frac{2sin^2x-1}{cosx-sinx}=\frac{2sin^2x-\left(sin^2x+cos^2x\right)}{cosx-sinx}=\frac{sin^2x-cos^2x}{cosx-sinx}\)
\(=\frac{\left(sinx-cosx\right)\left(sinx+cosx\right)}{cosx-sinx}=-\left(sinx+cosx\right)\)
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
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
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:
\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)
\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)
\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)
\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)
\(\left(sinx+siny\right)^2=3\Leftrightarrow sin^2x+sin^2y+2sinxsiny=3\) (1)
\(\left(cosx-cosy\right)^2=1\Leftrightarrow cos^2x+cos^2y-2cosx.cosy=1\) (2)
Cộng vế với vế (1) và (2):
\(sin^2x+cos^2x+sin^2y+cos^2y-2\left(cosx.cosy-sinx.siny\right)=4\)
\(\Leftrightarrow2-2cos\left(x+y\right)=4\)
\(\Rightarrow cos\left(x+y\right)=-1\)