\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{4-4sin^2x+sin^4x}+\sqrt{4-4cos^2x+cos^4x}\)
\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)
\(=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)\)
\(=3\)
Rút gọn \(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)
Rút gọn :
a) \(\cos\dfrac{x}{5}\cos\dfrac{2x}{5}\cos\dfrac{4x}{5}\cos\dfrac{8x}{5}\)
b) \(\sin\dfrac{x}{7}+2\sin\dfrac{3x}{7}+\sin\dfrac{5x}{7}\)
tính sin^6 + cos^6 - 2sin^4x - cos^4x + sin^2x
Cho tanx=-3, với \(\frac{\pi}{2}< x< \pi\). Tính giá trị của biểu thức
\(A=\frac{4sin^2x+5sinx.cosx+cos^2x}{sin^2x-2}\)
Rút gọn các biểu thức :
A= \(\sqrt{3}\) sin ( x - \(\dfrac{\pi}{3}\) ) + sin ( x + \(\dfrac{\pi}{6}\) )
B= cos7x cos5x - \(\sqrt{3}\) sin2x + sin7x sin5x
C= 2sin( 2x -\(\dfrac{\pi}{6}\)) + 4sin + 1
D= \(\sqrt{3}\) cos2x + sin2x + 2sin(2x - \(\dfrac{\pi}{6}\))
E= sin2x + 2\(\sqrt{2}\) cosx + 2sin(x + \(\dfrac{\pi}{4}\)) +3
Chứng minh
a) \(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{1}{2}cot^4x\)
b) \(\frac{cos2x}{cot^2x-tan^2x}=\frac{1}{4}sin^22x\)
CM BT ko phụ thuộc vào tham số x
\(A=2\left(cos^6x+sin^6x\right)-3\left(cos^4x+sin^4x\right)\)
B\(=\frac{tan^2x}{sin^2x.cos^2x}-\left(1+tan^2x\right)^2\)
Biết \(\dfrac{Sin^4x-Cos^4x+Cos^2x}{2\left(1-Cosx\right)}=Cos^k\dfrac{mx}{n}\).Trong đó \(\dfrac{m}{n}\) tối giản. Tìm m,k,n.
Rút gọn biểu thức sau:
A=4sinx*cosx*cos2x*cos4x
B=cos^4x -6cos^x*sin^2x+sim^4x
C=\(\frac{\text{cos2a-cos4a}}{sin4a+sin2a}\)
D=\(\frac{\text{cosa+cos3a+cos5a}}{sina+sin3a+sin5a}\)
E=sin^2(\(\frac{\pi}{8}\)+\(\frac{x}{2}\))-sin^2(\(\frac{\pi}{8}\)-\(\frac{x}{2}\))
F=\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
