\(cot2x-cot4x=\frac{cos2x}{sin2x}-\frac{cos4x}{sin4x}=\frac{sin4x.cos2x-cos4x.sin2x}{sin2x.sin4x}\)
\(=\frac{sin\left(4x-2x\right)}{sin2x.sin4x}=\frac{sin2x}{sin2x.sin4x}=\frac{1}{sin4x}\)
Đề bài sai
Rút gọn các biểu thức sau
1, \(\dfrac{1+\cot x}{1-\cot x}-\dfrac{2+2\cot^2x}{\left(\tan x-1\right)\left(\tan^2x+1\right)}\)
2, \(\sqrt{\sin^4x+6\cos^2x+3\cos^4x}+\sqrt{\cos^4x+6\sin^2x+3\sin^4x}\)
CMR :
a) \(\frac{sinx+sin3x+sin4x}{1+cosx+cos3x+cos4x}=tan2x\)
b) \(\frac{sin^22x+2cos\left(3\pi+2x\right)-2}{-3+4cos2x+cos\left(4x-\pi\right)}=\frac{1}{2}cot^4x\)
Chứng minh
a) \(\frac{1-sin2x}{1+sin2x}=cot^2\left(\frac{\pi}{4}-x\right)\)
b) \(\frac{Sin2x-2sinx}{sin2x+2sinx}=-tan^2\frac{x}{2}\)
CMR
\(\frac{cot^2\left(\frac{x}{2}\right)-cot^2\left(\frac{3x}{2}\right)}{cos^2\left(\frac{x}{2}\right).cosx.\left(1+cot^2\frac{3x}{2}\right)}=8\)
Đơn giản biểu thức : O = \(\frac{cot^2x-cos^2x}{cot^2x}+\frac{sinx.cosx}{cotx}\)
1. cos3a . sin a - sin3a . cos a =\(\frac{\sin4a}{4}\)
2. \(\frac{\cos^2x-\sin^2x}{\cot^2x-tan^2x}=\frac{1}{4}\sin^22x\)
3. \(\frac{\sin2x}{1+\cos2x}=tanx\)
4. rút gọn ; \(A=\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
chứng minh các đẳng thức sau
a) \(\tan^2x-\sin^2x=\tan^2x.\sin^2x\)
b) \(\tan x+\cot x=\frac{1}{\sin x.\cot x}\)
c) \(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
d) \(\frac{1}{1+\tan x}+\frac{1}{1+\cot x}=1\)
e) \(\left(1-\frac{1}{\cos x}\right)\left(1+\frac{1}{\cos x}\right)+\tan^2x=0\)
Chứng minh trong mọi tam giác ABC ta đều có :
a) \(\tan\frac{A}{2}.\tan\frac{B}{2}+\tan\frac{B}{2}.\tan\frac{C}{2}+\tan\frac{C}{2}.\tan\frac{A}{2}=1\)
b) \(\cot A.\cot B+\cot B.\cot C+\cot C.\cot A=1\)
chứng minh rằng
3) \(\frac{sin2x-sinx}{1-cosx+cos2x}=tanx\)
4) \(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+sin^{2014}x.tan^{2014}x}\)
