Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
$A=\frac{2\cos \frac{2x+y}{2}\sin \frac{x}{2}}{2\sin \frac{2x+y}{2}.\cos \frac{x}{2}}-\frac{2\cos \frac{2x+y}{2}\cos \frac{x}{2}}{-2\sin \frac{2x+y}{2}\sin \frac{x}{2}}$
$=\tan \frac{x}{2}.\cot \frac{2x+y}{2}+\cot \frac{x}{2}.\cot \frac{2x+y}{2}=\cot \frac{2x+y}{2}(\tan \frac{x}{2}+\cot \frac{x}{2})$
\(cosa=cos\left(2\pi-a\right)\)
Mà \(t=2\pi-\left(x+y+z\right)\) nên...
tư mã chiêu
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
Đặt \(\frac{x}{2}=a\)
\(\frac{sin2a+sina}{1+cos2a+cosa}=\frac{2sina.cosa+sina}{1+2cos^2a-1+cosa}=\frac{sina\left(2cosa+1\right)}{cosa\left(2cosa+1\right)}=\frac{sina}{cosa}=tana=tan\frac{x}{2}\)
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
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
\(x+2y=\dfrac{\pi}{2}\)
\(\Leftrightarrow x+y=\dfrac{\pi}{2}-y\) 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: \(A=\dfrac{siny-cosy}{siny+cosy}=\dfrac{\sqrt{2}sin\left(y-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(y-\dfrac{\pi}{4}\right)}=tan\left(y-\dfrac{\pi}{4}\right)\)