TÍNH
\(\sin^6\alpha+\cos^6\alpha+3\times\sin^2\alpha\times\cos^2\alpha\)
TT
Những câu hỏi liên quan
\(2(\sin\alpha-\cos\alpha)^2-(\sin\alpha+\cos\alpha)^2+6\sin\alpha\times\cos\alpha\)
CMR:\(sin^6\alpha+cos^6\alpha=1-3\times sin^2\alpha\times cos^2\alpha\)
\(\sin^6a+cos^6a=\left(sin^2a\right)^3+\left(cos^2a\right)^3\)=\(\left(sin^2a+cos^2a\right)\left(sin^4a-sin^2a\cdot cos^2a+cos^4a\right)\)
ma \(sin^2a+cos^2a=1\) nên ta có
=\(1\left(sin^4+cos^4-cos^2a\cdot sin^2a\right)\)
ma \(\left(sin^2a+cos^2a\right)^2=sin^4a+cos^4a+2sin^2a\cdot cos^2a\)
\(\Rightarrow sin^4a+cos^4a=\left(\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a\right)\)=\(\left(1^2-2sin^2a.cos^2a\right)\)
thay vao tren ta co
\(sin^6a+cos^6a=1\left(1-2sin^2a.cos^2a-sin^2a.cos^2a=1-3sin^2cos^2a\right)\)
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VT= \(\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)\)
\(=1.\left[\left(sin^2\alpha+cos^2\alpha\right)^2-3sin^2\alpha.cos^2\alpha\right]\)
\(=1-3sin^2\alpha.cos^2\alpha\)=VP
=>dpcm
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\(sin^6a+cos^6a\) \(=\left(\sin^2a\right)^3+\left(\cos^2a\right)^3=\left(\sin^2a+\cos^2a\right)\left(\sin^4a-\sin^2a\cdot\cos^2a+\cos^4a\right)\)
xomg rùi làm tương tự như mimh trieu nhé ^.^
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Xem thêm câu trả lời
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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cho tan\(\alpha\)=3/4. tinh
A=\(\dfrac{sin^3\alpha+cos^3\alpha}{2\sin\alpha\times\cos^2\alpha+\cos\alpha\times\sin^2\alpha}\)
ta có : \(A=\dfrac{sin^3\alpha+cos^3\alpha}{2sin\alpha.cos^2\alpha+cos^2\alpha.sin^2\alpha}\)
\(\Leftrightarrow A=\dfrac{\dfrac{sin^3\alpha}{cos^3\alpha}+\dfrac{cos^3\alpha}{cos^3\alpha}}{\dfrac{2sin\alpha.cos^2\alpha}{cos^3\alpha}+\dfrac{cos\alpha.sin^2\alpha}{cos^3\alpha}}=\dfrac{tan^3\alpha+1}{2tan\alpha+tan^2\alpha}\)
\(\Leftrightarrow A=\dfrac{\left(\dfrac{3}{4}\right)^3+1}{2\left(\dfrac{3}{4}\right)+\left(\dfrac{3}{4}\right)^2}=\dfrac{91}{132}\)
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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)\(\frac{sin^4\alpha-cos^4\alpha}{sin\alpha+cos\alpha}-sin\alpha+cos\alpha\)
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, \(A=\dfrac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha-\sin^2\alpha\right)}{\cos\alpha.\sin\alpha}\)
\(b,B=\sin^6\alpha+\cos^6\alpha+3\sin^6\alpha.\cos^2\alpha\)
\(\cos^2\alpha\times\cos^2\beta+\cos^2\alpha\times\sin^2\beta+\sin^2\alpha\)
\(\cos^2\alpha.\cos^2\beta+\cos^2\alpha.\sin^2\beta+\sin^2\alpha\)
\(=\cos^2\alpha.\left(\cos^2\beta+\sin^2\beta\right)+\sin^2\alpha\)
\(=\cos^2\alpha.1+\sin^2\alpha\)
\(=\cos^2\alpha+\sin^2\alpha\)
\(=1\)
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