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\(cotx-tanx-2tan2x=\frac{cosx}{sinx}-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}\)
\(=\frac{cos^2x-sin^2x}{\frac{1}{2}.2.sinxcosx}=\frac{cos2x}{\frac{1}{2}sin2x}=2\left(\frac{cos2x}{sin2x}-\frac{sin2x}{cos2x}\right)\)
\(=2\left(\frac{cos^22x-sin^22x}{\frac{1}{2}2sin2xcos2x}\right)=4\frac{cos4x}{sin4x}=4cot4x\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x-2.tanx.cotx+cot^2x=9\)
\(\Rightarrow tan^2x+cot^2x=11\)
\(\left(tanx+cotx\right)^2=tan^2x+cot^2x+2.tanx.cotx=11+2=13\)
\(\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=11\left(tanx+cotx\right)\left(tanx-cotx\right)=\pm33\sqrt{13}\)
\(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0;cosx< 0;tanx>0;cotx>0\)
\(tanx-3cotx=6\Leftrightarrow tanx-\frac{3}{tanx}=6\)
\(\Leftrightarrow tan^2x-6tanx-3=0\Rightarrow\left[{}\begin{matrix}tanx=3+2\sqrt{3}\\tanx=3-2\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)
\(\frac{1}{cos^2x}=1+tan^2x\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{-1}{\sqrt{1+tan^2x}}\) (do \(cosx< 0\))
\(\Rightarrow cosx=\frac{-1}{\sqrt{22+12\sqrt{3}}}\Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{\frac{15+6\sqrt{3}}{26}}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3+2\sqrt{3}}\)
Số xấu dữ dội, bạn tự thay vào kết quả :(
\(\left(tanx+cotx\right)^2=16\Leftrightarrow tan^2x+cot^2x+2=16\Rightarrow tan^2x+cot^2x=14\)
\(A=tan^2x+4cot^2x+4+4tan^2x+cot^2x+4\)
\(A=5\left(tan^2x+cot^2x\right)+8=5.14+8=78\)
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
Lời giải:
a)
\(\frac{\cos (a-b)}{\cos (a+b)}=\frac{\cos a\cos b+\sin a\sin b}{\cos a\cos b-\sin a\sin b}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\)
b)
\(2(\sin ^6a+\cos ^6a)+1=2(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=2(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^4a+\cos ^4a+2\sin ^2a\cos ^2a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^2a+\cos ^2a)^2+1\)
\(=3(\sin ^4a+\cos ^4a)-1^2+1=3(\sin ^4a+\cos ^4a)\)
c)
\(\frac{\tan a-\tan b}{cot b-\cot a}=\frac{\tan a-\tan b}{\frac{1}{\tan b}-\frac{1}{\tan a}}\) (nhớ rằng \(\tan x.\cot x=1\rightarrow \cot x=\frac{1}{\tan x}\) )
\(=\frac{\tan a-\tan b}{\frac{\tan a-\tan b}{\tan a\tan b}}=\tan a\tan b\)
d)
\((\cot x+\tan x)^2-(\cot x-\tan x)^2=(\cot ^2x+\tan ^2x+2\cot x\tan x)-(\cot ^2x-2\cot x\tan x+\tan ^2x)\)
\(=4\cot x\tan x=4.1=4\)
e)
\(\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)
\(=\sin ^2a-\sin a\cos a+\cos ^2a=(\sin ^2a+\cos ^2a)-\sin a\cos a=1-\sin a\cos a\)
Vậy ta có đpcm.
\(2sin\left(\frac{\pi}{4}+a\right)sin\left(\frac{\pi}{4}-a\right)=cos2a-cos\left(\frac{\pi}{2}\right)=cos2a\)
\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=-\frac{2\left(cos^2x-sin^2x\right)}{2sinx.cosx}=\frac{2cos2x}{sin2x}=-2cot2x=-\frac{2}{tan2x}\)
Lời giải:
a)
\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)
Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$
$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$
$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$
$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$
-------------------------
$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$
$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$
$\Rightarrow \cos 2a=\frac{-7}{25}$
$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$
$\Rightarrow \cot 2a=\frac{-7}{24}$
\(\left(tanx+cotx\right)^2=m^2\)
\(\Leftrightarrow tan^2x+cot^2x+2=m^2\)
\(\Leftrightarrow tan^2x+cot^2x=m^2-2\)
\(\Rightarrow\left(tan^2x+cot^2x\right)^2=\left(m^2-2\right)^2\)
\(\Leftrightarrow tan^4x+cot^4x+2=m^4-4m^2+4\)
\(\Leftrightarrow tan^4x+cot^4x=m^4-4m^2+2\)
\(\Rightarrow a+b+c+d+e=1+0-4+0+2=-1\)
\(A=\frac{cosx}{sinx}-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{sin4x}-\frac{8sin8x}{cos8x}\)
\(A=\frac{cos^2x-sin^2x}{sinx.cosx}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{2cos2x}{sin2x}-\frac{2sin2x}{cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{2cos^22x-2sin^22x}{sin2x.cos2x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}\)
\(A=\frac{4cos4x}{sin4x}-\frac{4sin4x}{cos4x}-\frac{8sin8x}{8cos8x}=\frac{8cos8x}{sin8x}-\frac{8sin8x}{cos8x}\)
\(A=\frac{16cos16x}{sin16x}=16cot16x\)
\(B=\frac{1}{2}.2sinx.cosx.cos2x.cos4x.cos8x\)
\(B=\frac{1}{2}sin2x.cos2x.cos4x.cos8x\)
\(B=\frac{1}{4}sin4x.cos4x.cos8x\)
\(B=\frac{1}{8}sin8x.cos8x\)
\(B=\frac{1}{16}sin16x\)