\(2(\sin\alpha-\cos\alpha)^2-(\sin\alpha+\cos\alpha)^2+6\sin\alpha\times\cos\alpha\)
TL
Những câu hỏi liên quan
Cho \(\tan\alpha=\dfrac{3}{5}\). Tính giá trị của các biểu thức sau:
M=\(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
N=\(\dfrac{\sin\alpha\times\cos\alpha}{\sin^2\alpha-\cos^2\alpha}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
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CMR: \(\frac{\sin^4\alpha-\cos^2\alpha+2\cos^4\alpha-\cos^6\alpha}{\cos^4\alpha-\sin^2\alpha+2\sin^4\alpha-\sin^6\alpha}=\tan^6\alpha\)
\(\frac{\sin^4\alpha-\cos^2\alpha+2\cos^4\alpha-\cos^6\alpha}{\cos^4\alpha-\sin^2\alpha+2\sin^4\alpha-\sin^6\alpha}=\frac{\sin^4\alpha-\cos^2\alpha\left(1-\cos^2\alpha\right)^2}{\cos^4\alpha-\sin^2\alpha\left(1-\sin^2\alpha\right)^2}\)
\(=\tan^4\alpha.\frac{1-\cos^2\alpha}{1-\sin^2\alpha}=\tan^6\alpha\)
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tính :
\(E=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cdot\cos^2\alpha\)
\(F=3\sin^3\alpha+\cos^3\alpha-2\sin^6\alpha+\cos^6\alpha\)
\(G=\sqrt{\sin^4\alpha+4\cos^2\alpha}+\sqrt{\cos^4\alpha+4\sin^2\alpha}\)
E = sin^6 + cos^6 + 3sin^2.cos^2
= (sin^2 + cos^2)(sin^4 - sin^2.cos^2 + cos^4) + 3 sin^2.cos^2
= (sin^2 + cos^2)^2 - 3sin^2.cos^2 + 3sin^2.cos^2
= 1
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tính
a) \(\tan^2\alpha-\sin^2\alpha-\tan^2\alpha\times\sin^2\alpha\)
b)\(\frac{sin^4\alpha-cos^4\alpha}{sin\alpha+cos\alpha}-sin\alpha+cos\alpha\)
\(B=\dfrac{\sin^2\alpha-2.\sin\alpha.\cos\alpha+3.\cos^2\alpha}{2.\sin^2\alpha+\sin\alpha.\cos\alpha-2\cos^2\alpha}\)
Sin² α+ cos^4 α + 2sin α . cos^2 α
Sin^6 α – sin^6 α + 3sin α . Cos^2 α
Chứng minh các biểu thức sau không phụ thuộc vào các góc nhọn alphaa) Ccos^4alpha+sin^2alpha.cos^2alpha+sin^2alphab) Dsin^2alpha.sin^2beta+sin^2alpha.cos^2beta+cos^2alphac) Esin^6alpha+sin^6beta+3.sin^2alpha.cos^2alpha d) Mfrac{left(cosalpha-sinalpharight)^2-left(cosalpha+sinalpharight)^2}{cosalpha.sinalpha}
Đọc tiếp
Chứng minh các biểu thức sau không phụ thuộc vào các góc nhọn \(\alpha\)
a) \(C=\cos^4\alpha+\sin^2\alpha.\cos^2\alpha+\sin^2\alpha\)
b) \(D=\sin^2\alpha.\sin^2\beta+\sin^2\alpha.\cos^2\beta+\cos^2\alpha\)
c) E=\(\sin^6\alpha+\sin^6\beta+3.\sin^2\alpha.\cos^2\alpha\)
d) \(M=\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha+\sin\alpha\right)^2}{\cos\alpha.\sin\alpha}\)
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,\(C=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)
RÚT GỌN:
a, \(A=\dfrac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha-\sin^2\alpha\right)}{\cos\alpha.\sin\alpha}\)
\(b,B=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)