\(A=cosx+2cos^2x+8cos^3x-6cosx\)
\(=cosx\left(8cos^2x+2cosx-5\right)\)
\(B=\left(sin^3x-2sin^5x\right)-\left(2cos^5x-cos^3x\right)\)
\(=sin^3x\left(1-2sin^2x\right)-cos^3x\left(2cos^2x-1\right)\)
\(=sin^3x.cos2x-cos^3x.cos2x\)
\(=cos2x\left(sin^3x-cos^3x\right)\)
\(=cos2x\left(sinx-cosx\right)\left(1+sinx.cosx\right)\)
Cm biểu thức ko phụ thuộc x
B=\(\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)\left(1+cosx\right)}\)
Cm
\(\dfrac{1+sin2x-cos2x}{1+sin2x+cos1x}=tanx\)
Tính giá trị cot8x. Cm biểu thức không phụ thuộc x
\(\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)\left(1+cosx\right)}\)
Cm \(\dfrac{1+sin2x-cos2x}{1+sin2x+cos2x}\)
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) \(\cos x\cos\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{3}+x\right)=\frac{1}{4}\cos3x\)
b) \(\sin5x-2\sin x\left(\cos4x+\cos2x\right)=\sin x\)
Chứng minh rằng: (Pls help me)
a, \(\frac{1}{\sin x}+\cot x=\cot\frac{x}{2}\)
b, \(\frac{1-\cos x}{\sin x}=\tan\frac{x}{2}\)
c,\(\tan\frac{x}{2}\left(\frac{1}{\cos x}+1\right)=\tan x\)
d,\(\frac{\sin2a}{2\cos a\left(1+\cos a\right)}=\tan\frac{a}{2}\)
e,\(\cot x+\tan\frac{x}{2}=\frac{1}{\sin x}\)
f,\(3-4\cos2x+\cos4x=8\sin^4x\)
g,\(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
h,\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
i,\(\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\)
l,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=2\left(cos^6x+sin^6x\right)-3\left(cos^4x+sin^4x\right)\)
b) \(B=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)^2-sin^8x-cos^8x\)
c) \(C=\dfrac{sin^2x}{1+cotgx}+\dfrac{cos^2x}{1+tgx}+sinx.cosx\)
d) \(D=\dfrac{cotg^2a-cos^2x}{cotg^2x}+\dfrac{sinx.cosx}{cotgx}\)
e) \(E=3\left(sin^8x-cos^8x\right)+4\left(cos^6x-2sin^6x\right)+6sin^4x\)
f) \(F=\dfrac{tg^2x}{sin^2x.cos^2x}-\left(1+tg^2x\right)^2\)
Rut gon
A=\(\frac{sin\left(x+y\right)-sinx}{sin\left(x+y\right)+sinx}\) - \(\frac{cos\left(x+y\right)+cosx}{cos\left(x-y\right)-cosx}\)
chứng minh rằng
1) \(tanx=\frac{1-cos2x}{sin2x}\)
2)\(\frac{sin\left(60^0-x\right).cos\left(30^{0^{ }}-x\right)+cos\left(60^{0^{ }}-x\right).sin\left(30^{0^{ }}-x\right)}{sin4x}=\frac{1}{2sin2x}\)
3) \(4cos\left(60^0+a\right).cos\left(60^0-a\right)+2sin^2a=cos2a\)
Rút gọn các biểu thức sau:
a, \(A=\sin^2\left(a-b\right)+\sin^2b+2\sin\left(a-b\right).\sin b.\cos a\)
b, \(B=\cos^2a+\cos^2\left(a+b\right)-2\cos a.\cos b.\cos\left(a+b\right)\)
Mọi người giúp mình với ạ!!!
