ta có : \(A=sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x\right)^3+\left(cos^2x\right)^3+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin^2x.cos^2x\)
\(=1^3-3sin^2x.cos^2x+3sin^2x.cos^2x=1\)
rút gọn biểu thức :
A = 1 + \(\dfrac{2\sin\alpha.\cos\alpha}{\cos^2\alpha-\sin^2\alpha}\)
B = \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
Rút gọn các biểu thức:
a)\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)\
b) \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
Chứng minh:
sin^4-sin^6+cos^4-cos^6=sin^2.cos^2
Rút gọn
\(1,D=cos^220^0+cos^230^0+cos^240^0+cos^250^0+cos^260^0+cos^270^0\)
\(2,E=sin^25^0+sin^225^0+sin^245^0+sin^265^0+sin^285^0\)
\(3,F=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)
Sin² α+ cos^4 α + 2sin α . cos^2 α
Sin^6 α – sin^6 α + 3sin α . Cos^2 α
\(K=\sin^6\alpha+\cos^6\beta+3.\sin^2\alpha.\cos^2\alpha\)
Chứng minh:
a)\(cot^2\alpha-cos^2\alpha\cdot cot^2\alpha=cos^2\alpha\)
b)\(tan^2\alpha-sin^2\alpha\cdot tan^2\alpha=sin^2\alpha\)
c) \(\dfrac{1-cos^2}{sin\alpha}\) = \(\dfrac{sin\alpha}{1+cos\alpha}\)
d)\(tan^2\alpha-sin^2\alpha=tan^2\cdot sin^2\alpha\)
e) \(\sin^6\alpha+cos^6\alpha+3sin^2\cdot cos^2\alpha=1\)
. Chứng minh rằng các hệ thức sau không phụ thuộc vào anfa và bêta
A= (sin a +cos a )2 - (sin a - cos a)2 ( Oo < a < 9Oo )
B= sin4 a + cos 4 a + 2 sin2 a cos 2 a ( Oo < a < 9Oo )
C= Cos4 a + sin2 a cos2 a + sin2 a. ( Oo < a < 9Oo )
D= Sin2 a Sin2 p + sin2 a cos2 p + cos 2 a ( Oo < a ; p < 9Oo )
E= Sin 6 a + cos 6 a + 3 sin2 a cos2 a ( Oo < a < 9Oo )
p/s : a là anfa , p là pêta
THANKS
TÍNH
\(\sin^6\alpha+\cos^6\alpha+3\times\sin^2\alpha\times\cos^2\alpha\)
