\(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{\left(1-cos^2x\right)^2+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{1-cos^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{sin^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=2\sqrt{sin^2x+cos^4x}\)
\(\sqrt{sin^4x+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{\left(1-cos^2x\right)^2+cos^2x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{1-cos^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=\sqrt{sin^2x+cos^4x}+\sqrt{sin^2x+cos^4x}\)
\(=2\sqrt{sin^2x+cos^4x}\)
Rút gọn biểu thức \(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4sin^2x}\) .
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
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.
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
Bài 1: chứng minh rằng
a, \(\dfrac{\sin x+\cos x-1}{1-\cos x}\)=\(\dfrac{2\cos x}{\sin x-\cos x+1}\)
b, \(\cot^2x-\cos^2x=\cot^2x\cos^2x\)
chứng minh biểu thức sau không phụ thuộc vào x:
\(\dfrac{sin^6x+cos^6x+2}{sin^4x+cos^6x+1}\)
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=\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)