a, Ta có :
\(\dfrac{2sin4x+\sqrt{2}}{2}\)=sin4x + \(\dfrac{\sqrt{2}}{2}\)=sin4x + sin\(\dfrac{\pi}{4}\)=2sin(4x+\(\dfrac{\pi}{4}\)).cos(4x-\(\dfrac{\pi}{4}\))
=> 2sin4x+\(\sqrt{2}\) = 4sin(4x+\(\dfrac{\pi}{4}\)).cos(4x-\(\dfrac{\pi}{4}\))
a, Ta có :
\(\dfrac{2sin4x+\sqrt{2}}{2}\)=sin4x + \(\dfrac{\sqrt{2}}{2}\)=sin4x + sin\(\dfrac{\pi}{4}\)=2sin(4x+\(\dfrac{\pi}{4}\)).cos(4x-\(\dfrac{\pi}{4}\))
=> 2sin4x+\(\sqrt{2}\) = 4sin(4x+\(\dfrac{\pi}{4}\)).cos(4x-\(\dfrac{\pi}{4}\))
Chứng minh:
1.\(\dfrac{\cot^2x-\sin^2x}{\cot^2x-\tan^2x}=\sin^2x\cdot\cos^2x\)
2.\(\dfrac{1-\sin x}{\cos x}-\dfrac{\cos x}{1+\sin x}=0\)
3.\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cot x}=\cos x\)
4.\(\dfrac{\tan x}{1-\tan^2x}\cdot\dfrac{\cot^2x-1}{\cot x}=1\)
5.\(\dfrac{1+\sin^2x}{1-\sin^2x}=1+2\tan^2x\)
biến đổi thành tích biểu thức
1. cos x + sin 2x - cos 3x
2. sin 3x - sin x +sin 2x
chứng minh các đẳng thức sau : a) \(\frac{1+2sinxcosx}{sin^2x-cos^2x}\) = \(\frac{tan+1}{tan-1}\) ; b) sin4x - cos4x = 1 - 2cos2x ; c) sin4x + cos4x = \(\frac{3}{4}\) + \(\frac{1}{4}\)cosx ; d) sin6x + cos6x = \(\frac{5}{8}\) + \(\frac{3}{8}\)cos4x ; e) cotx - tanx = 2cot2x ; f) \(\frac{sin2x+sin4x+sin6x}{1+cos2x+cos4x}\) = 2sin2x
1. Cho \(2\cos\left(\alpha+\beta\right)=\cos\alpha\cos\left(\pi+\beta\right)\)
Tính \(A=\dfrac{1}{2\sin^2\alpha+3\cos^2\alpha}+\dfrac{1}{2\sin^2\beta+3\cos^2\beta}\)
2. Rút gọn: a) \(A=4\cos\dfrac{2x}{3}\cos\dfrac{\pi+2x}{3}\cos\dfrac{\pi-2x}{3}\)
b) \(B=\dfrac{\sin\left(a-b\right).\sin\left(a+b\right)}{\cos^2a.\sin^2b}-\tan^2a.\cot^2b\)
3. Chứng minh rằng: Nếu \(2\tan a=\tan\left(a+b\right)\) thì:
a) \(\sin b=\sin a.\cos\left(a+b\right)\)
b) \(3\sin b=\sin\left(2a+b\right)\)
Cho góc x với cos=-1/2. Tính giá trị biểu thức B=cos^2x+sin^2x+tan^2x
chứng minh các đẳng thức sau :
a) \(\frac{1+2\sin x\cos x}{\sin^2x-\cos^2x}\)=\(\frac{\tan x+1}{\tan x-1}\)
b) \(\sin\)4x + \(\cos\)4x =\(\frac{3}{4}\)+\(\frac{1}{4}\)\(\cos\)x
c) \(\sin\)6x + \(\cos\)6x = \(\frac{5}{8}\) + \(\frac{1}{8}\)\(\cos\)4x
d) \(\cot\)x - \(\tan\)x = 2\(\cot\)2x
Chứng minh các biểu thức sau không phụ thuộc vào x:
1, \(A=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
2, \(B=cos^6x+2sin^4x.cos^2x+3sin^2x.cos^4x+sin^4x\)
3, \(C=cos\left(x-\dfrac{\pi}{3}\right).cos\left(x+\dfrac{\pi}{4}\right)+cos\left(x+\dfrac{\pi}{6}\right).cos\left(x+\dfrac{3\pi}{4}\right)\)
4, \(D=cos^2x+cos^2\left(x+\dfrac{2\pi}{3}\right)+cos^2\left(\dfrac{2\pi}{3}-x\right)\)
5, \(E=2\left(sin^4x+cos^4x+sin^2x.cos^2x\right)-\left(sin^8x+cos^8x\right)\)
6, \(F=cos\left(\pi-x\right)+sin\left(\dfrac{-3\pi}{2}+x\right)-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\dfrac{3\pi}{2}-x\right)\)
Chứng minh đẳng thức:
1 ,\(tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)+cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)=\dfrac{2}{cosx}\)
2 ,\(sin^8x-cos^8x=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)\)
3 ,\(3-4cos2x+cos4x=8sin^4x\)
4 ,\(sin\left(2x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)-cos\left(2x+\dfrac{\pi}{3}\right).cos\left(\dfrac{2\pi}{3}-x\right)=cosx\)
5 ,\(\sqrt{3}cos2x+sin2x+sin\left(4x-\dfrac{\pi}{3}\right)=4cos\left(2x-\dfrac{\pi}{6}\right).sin^2\left(x+\dfrac{\pi}{6}\right)\)
Chứng minh các đẳng thức:
\(cos^3xsinx-sin^3xcosx=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\dfrac{1}{4}\left(3+cos4x\right)\)