Cái này mình vừa làm ban nãy rồi mà-.-
Ta có: \(2\sin^2\alpha+\cot^2\alpha\cdot\sin^2\alpha+\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)+\left(\sin^2\alpha+\frac{\cos^2\alpha}{\sin^2\alpha}\cdot\sin^2\alpha\right)\)
\(=1+\left(\sin^2\alpha+\cos^2\alpha\right)\)
\(=1+1=2\)
Rút gọn các biểu thức:
a)\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
b)\(\cot^2\alpha-\cos^2\alpha.\cot^2\alpha\)
c)\(\sin\alpha.\cos\alpha\left(\tan\alpha+\cot\alpha\right)\)
d)\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
Rút gọn
B= \(2\sin^2\alpha+\cot^2\alpha.\sin^2\alpha+\cos^2\alpha\)
rút gọn
a)A=\(\frac{1+2cos\alpha.sin\alpha}{cos^2\alpha-sin^2\alpha}\)
b)B=\(\left(1+\cot^2\alpha\right)\left(1-sin^2\alpha\right)\)-\(\left(1+\cot^2\alpha\right)\left(1-\cos^2\alpha\right)\)
c)C=\(\sin^6\alpha+\cos^6\alpha\)+\(3\sin^2\alpha.cos^2\alpha\)
Tinh ti so luong giac sau :
\(\sin\alpha\times\cos\alpha+\frac{\sin^2\alpha}{1+\cot\alpha}+\frac{\cos^2\alpha}{1+\tan\alpha}\)
rút gọn biểu thức sau:
b, \(\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha-\sin^2\alpha\right)}{\cos\alpha.\sin\alpha}\)
c,\(C=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)
Chứng minh rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của góc nhọn \(\alpha\)
a) A = \(\frac{\cot^2\alpha-\cos^2\alpha}{\cot^2\alpha}-\frac{\sin\alpha.\cos\alpha}{\cot\alpha}\)
b) B = \(\left(\cos\alpha-\sin\alpha\right)^2+\left(\cos\alpha+\sin\alpha\right)^2+\cos^4\alpha-\sin^4\alpha-2\cos^2\alpha\)
c) C = \(\sin^6x+\cos^6x+3\sin^2x.\cos^2x\)
1/ Rút gọn
\(A=\cot^2x\left(\cos^2x-1+\sin^2x\right)+\sin^2x\)
\(B=\cos^4\alpha-\sin^4\alpha+2\sin^2\alpha+8\)
2) Rút gọn
a)\(1-\sin^22\)
b)\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)\)
c)\(1+\sin^2\alpha+\cos^2\alpha\)
d)\(\sin\alpha-\sin\alpha.\cos^2\alpha\)
e)\(\sin^2\alpha+\cos^2\alpha+2\sin^2\alpha.\cos^2\alpha\)
f)\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha\)
g)\(\cos^2\alpha+\tan^2\alpha.\cos^2\alpha\)
h)\(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
CMR: \(\frac{\sin^2\alpha}{\cos\alpha\left(1+\tan\alpha\right)}-\frac{\cos^2\alpha}{\sin\alpha\left(1+\cot\alpha\right)}=\sin\alpha-\cos\alpha\)
