Ta có \(\sin^2x+\cos^2x=1\Rightarrow\cos^2x=1-\sin^2x\)
Từ dó \(A=2\left(1-\sin^2x\right)^2-\sin^4x+\sin^2x\left(1-\cos^2x\right)+3\sin^2x\)
\(=2\left(1-2\sin^2x+\sin^4x\right)-\sin^4x+\sin^2x\left(1-\sin^2x\right)+3\sin^2x\)
\(=2-4\sin^2x+2\sin^4x-\sin^4x+\sin^2x-\sin^4x+3\sin^2x=2\)
Vậy A=2
Chứng minh các biểu thức sau không phụ thuộc vào x:
a) \(A=\cos^4x-\sin^4x+2\sin^2x+\tan2x.\cot2x\)
b) \(B=\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
c) \(C=3\left(\sin^8x-\cos^8x\right)+4\left(\cos^6x-2\sin^6x\right)+6\sin^4x\)
d) \(D=2\left(\sin^4x+\cos^4x+\sin^2x.\cos^2x\right)-\left(\sin^8x+\cos^8x\right)\)
Chứng minh :
a \(\sin^4x+\cos^4x=1-2\sin^2x.\cos^2x\)
b.\(\sin^6x+\cos^6x=1-3\sin^2x.\cos^2x\)
giá trị biểu thức
A=\(\cos^4x+2\sin^4x\cos^2x+\sin^4x+2\cos^4x\sin^2x+1\)
Chứng minh: \(\sin^4x+\cos^2x\cdot\sin^2x+\sin^2x=2\sin^2x\)
Chứng minh
a)\(\left(\sin x+\cos x\right)^2=1+2\sin x\)\(\cos x\)
b)\(\left(\sin x+\cos x\right)^2+\left(\sin x-\cos x\right)^2=2\)
c)\(\sin^4x+\cos^4x=1-2\sin^2x\cos^2x\)
\(\cos^4x+\sin^2x.\cos^2x+\sin^2x\)
A=sin^6x +3sin^4x nhân cos^2x+3sin^2x nhân cos^4x+cos^6x
Thu gọn:
a/ cot^2x-cos^2x-cot^2x.cos^2x
b/ (sin^4x+cos^4x-1).(tan^2x+cot^2x+2)
\(\cos^4x+\sin^2x.cos^2x+sin^2x\)
