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sin x+cosx=m
=>(sinx+cosx)^2=m^2
=>1+2*cosx*sinx=m^2
=>2*sinx*cosx=m^2-1
=>\(sinx\cdot cosx=\dfrac{m^2-1}{2}\)
\(sin^3x+cos^3x=\left(sinx+cosx\right)^3-3\cdot sinx\cdot cosx\cdot\left(sinx+cosx\right)\)\(=m^3-3\cdot\dfrac{m^2-1}{2}\cdot m\)
\(=m^3-\dfrac{3m^3-3m}{2}\)
\(=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{-m^3+3m}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{3sinx-4cosx}{cosx+2sinx}=\frac{\frac{3sinx}{cosx}-4}{1+\frac{2sinx}{cosx}}=\frac{3tanx-4}{1+2tanx}=\frac{3.5-4}{1+2.5}=...\)
\(B=\frac{\frac{sinx}{cos^3x}+\frac{sin^3x}{cos^3x}}{\frac{3cos^3x}{cos^3x}+\frac{cosx}{cos^3x}}=\frac{tanx.\frac{1}{cos^2x}+tan^3x}{3+\frac{1}{cos^2x}}=\frac{tanx\left(1+tan^2x\right)+tan^3x}{3+\left(1+tan^2x\right)}=\frac{5\left(1+5^2\right)+5^3}{3+1+5^2}=...\)
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Lời giải:
a)
\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)
Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$
$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$
$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$
$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$
-------------------------
$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$
$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$
$\Rightarrow \cos 2a=\frac{-7}{25}$
$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$
$\Rightarrow \cot 2a=\frac{-7}{24}$
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\(A=sin^3x\cdot\left(1+\dfrac{cosx}{sinx}\right)+cos^3x\left(1+\dfrac{sinx}{cosx}\right)\)
\(=sin^2x\left(sinx+cosx\right)+cos^2x\left(cosx+sinx\right)\)
=cosx+sinx
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\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giả sử biểu thức xác định:
\(\dfrac{tanx-sinx}{sin^3x}=\dfrac{\dfrac{sinx}{cosx}-sinx}{sin^3x}=\dfrac{sinx-cosxsinx}{cosxsin^3x}\)
\(=\dfrac{sinx\left(1-cosx\right)}{sin^3xcosx}\)\(=\dfrac{1-cosx}{cosxsin^2x}=\dfrac{1-cosx}{cosx\left(1-cos^2x\right)}=\dfrac{1}{cosx\left(1+cosx\right)}\).