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\(\sin^6x+\cos^6x\\ =\left(\sin^2x\right)^3+\left(\cos^2x\right)^3\\ =\left(\sin^2x+\cos^2x\right)^3-3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)\\ =1-3\sin^2x\cos^2x\left(đpcm\right)\)
\(sin^6x+cos^6x\)
=\(\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
=\(sin^4x-sin^2x.cos^2x+cos^4x\)
=\(\left(1-2sin^2x.cos^2x\right)-sin^2x.cos^2x\)
=\(1-3sin^2x.cos^2x\)(đpcm)
➞\(sin^6x+cos^6x\)=\(1-3sin^2x.cos^2x\)
Có \(\sin^2x+\cos^2x=1\Rightarrow\sin^2x-\cos^2x=1-2\cos^2x\)
\(\Rightarrow VT=\frac{\sin^2x-\cos^2x}{\sin^2x.\cos^2x}=\frac{\sin^4x-\cos^4x}{\sin^2x.\cos^2x}=\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=\tan^2x-\cot^2x=VP\)
\(VP=\frac{2\sin^2x-1}{\sin^4x}=\frac{\sin^2x+\sin^2x-1}{\sin^4x}=\frac{\sin^2x-\cos^2x}{\sin^4x}\)
\(=\frac{\left(\sin^2x-\cos^2x\right).1}{\sin^4x}=\frac{\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)}{\sin^4x}=\frac{\sin^4x-\cos^4x}{\sin^4x}\)
\(=1-\cot^4x\)=VT
Lời giải:
Bạn chú ý lần sau gõ đề bài cho chuẩn xác. Không có dấu ngoặc () rất dễ gây lầm đề.
\(\cos ^4x(2\cos ^2x-3)+\sin ^4x(2\sin ^2x-3)\)
\(=2\cos ^6x-3\cos ^4x+2\sin ^6x-3\sin ^4x\)
\(=2[(\cos ^2x)^3+(\sin ^2x)^3]-3(\cos ^4x+\sin ^4x)\)
\(=2(\cos^2x+\sin ^2x)(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)
\(=2(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)
\(=-(\cos ^4x+2\cos ^2x\sin ^2x+\sin ^4x)=-(\cos ^2x+\sin ^2x)^2\)
\(=-1^2=-1\) là giá trị không phụ thuộc vào $x$. (đpcm)
Có \(\sin^2x+\cos^2x=1\Rightarrow2\sin^2x=1-\cos^2x+\sin^2x\)
\(\Rightarrow1+\sin^2x=2\sin^2x+\cos^2x\)
\(\Rightarrow VT=\frac{2\sin^2x+\cos^2x}{\cos^2x}=2\tan^2x+1\)
c)
\(\cos\left(x\right)^4+\sin\left(x\right)^2\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)\cos\left(x\right)^2+\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)
\(\cos\left(x\right)^4-\sin\left(x\right)^4+2\sin\left(x\right)^2\\ =\left(\cos\left(x\right)^2-\sin\left(x\right)^2\right)\left(\cos\left(x\right)^2+\sin\left(x\right)^2\right)+2\sin\left(x\right)^2\\ =\cos\left(2x\right)\cdot1+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2-\sin\left(x\right)^2+2\sin\left(x\right)^2\\ =\cos\left(x\right)^2+\sin\left(x\right)^2\\ =1\)
\(\cos^4x-\sin^4x=\left(\cos^2x-\sin^2x\right)\left(\cos^2x+\sin^2x\right)\)
\(=\cos^2x-\sin^2x=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1\)
(đpcm)