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NV
30 tháng 10 2019

\(A=\sqrt{\left(1-cos^2x\right)^2+4cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4sin^2x}\)

\(=\sqrt{cos^4x+2cos^2x+1}+\sqrt{sin^4x+2sin^2x+1}\)

\(=\sqrt{\left(cos^2x+1\right)^2}+\sqrt{\left(sin^2x+1\right)^2}\)

\(=sin^2x+cos^2x+2=3\)

b/

\(3\left(sin^8x-cos^8x\right)=3\left(sin^4x+cos^4x\right)\left(sin^4x-cos^4x\right)\)

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)\)

\(=3sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x-3cos^6x\)

\(\Rightarrow B=-5sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x+cos^6x+6sin^4x\)

\(=-5sin^6x-3sin^4x\left(1-sin^2x\right)+3cos^4x\left(1-cos^2x\right)+cos^6x+6sin^4x\)

\(=-2sin^6x-2cos^6x+3sin^4x+3cos^4x\)

\(=-2\left(1-3sin^2x.cos^2x\right)+3\left(1-2sin^2x.cos^2x\right)\)

\(=-2+3=1\)

NV
10 tháng 4 2019

\(\left(sin^4x+cos^4x+cos^2x.sin^2x\right)^2-sin^8x\)

\(=\left(sin^4x+cos^2x\left(cos^2x+sin^2x\right)\right)^2-sin^8x\)

\(=\left(sin^4x+cos^2x\right)^2-sin^8x=\left(sin^4x+cos^2x-sin^4x\right)\left(sin^4x+cos^2x+sin^4x\right)\)

\(=cos^2x\left(2sin^4x+cos^2x\right)=2sin^4x.cos^2x+cos^4x\)

Tương tự: \(\left(sin^4x+cos^4x+sin^2xcos^2x\right)^2-cos^8x\)

\(=\left(cos^4x+sin^2x\left(sin^2x+cos^2x\right)\right)^2-cos^8x\)

\(=\left(cos^4x+sin^2x\right)^2-cos^8x\)

\(=\left(cos^4x+sin^2x-cos^4x\right)\left(cos^4x+sin^2x+cos^4x\right)\)

\(=sin^2x\left(2cos^4x+sin^2x\right)=2sin^2x.cos^4x+sin^4x\)

\(\Rightarrow M=2sin^2x.cos^4x+2sin^2x.cos^2x+sin^2x+cos^4x\)

\(M=2sin^2x.cos^2x\left(cos^2x+sin^2x\right)+sin^4x+cos^4x\)

\(M=2sin^2x.cos^2x+sin^4x+cos^4x\)

\(M=\left(sin^2x+cos^2x\right)^2=1\)

30 tháng 3 2017

a)\(\eqalign{ & A\sin {x \over 5} = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} = {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over {16}}\sin {{16x} \over 5} \cr} \)

Suy ra biểu thức rút gọn \(A =\sin{{16x} \over 5}:16\sin {x \over 5}\)

b)\(\eqalign{ & B = \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7} = 2\sin {{3x} \over 7} + (\sin {x \over 7} + \sin {{5x} \over 7}) \cr & = 2\sin {{3x} \over 7} + 2\sin {1 \over 2}({{5x} \over 7} + {x \over 7})cos{1 \over 2}({{5x} \over 7} - {x \over 7}) \cr & = 2\sin {{3x} \over 7}(1 + \cos {{2x} \over 7}) = 4\sin {{3x} \over 7}{\cos ^2}{x \over 7} \cr}\)

NV
10 tháng 4 2021

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-8sin^6x+6sin^4x\)

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6sin^4x\left(1-sin^2x\right)\)

\(=sin^6x+3sin^4x.cos^2x+3cos^2x.sin^4x+cos^6x\)

\(=\left(sin^2x+cos^2x\right)^3=1\)