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7 tháng 9 2016

\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)

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

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

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

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

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

\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)

29 tháng 10 2018

\(\sin^6a+\cos^6a+3\sin^2a-\cos^2a\\ =\sin^6a+3\sin^2\cos^2\left(\sin^2a+\cos^2a\right)+\cos^6a-3\sin^2a\cos^2a\left(\sin^2a+\cos^2a\right)+3\sin^2a-\cos^2a\\ =\left(\sin^2a+\cos^2a\right)^3-3\sin^2a.\cos^2a.1+3\sin^2a-cos^2a\\ =1^3-\cos^2a+3\sin^2a-3\sin^2\cos^2\\ =\left(1-\cos^2a\right)\left(3\sin^2a+1\right)\)

9 tháng 11 2017

Đặt \(\sin^2\alpha=x\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)

\(A=x^3+\left(1-x\right)^3+3x-\left(1-x\right)=x^3+1-3x+3x^2-x^3+3x-1+x=3x^2+x\)

Vậy \(A=3\sin^4\alpha+\sin^2\alpha\). NHỚ NHA!

25 tháng 7 2023

\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)

b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)

4 tháng 8 2018

a) ta có : \(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

\(\Leftrightarrow A=sin^2\alpha+2sin\alpha.cos\alpha+cos^2\alpha+sin^2\alpha-2sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow A=2\left(sin^2\alpha+cos^2\alpha\right)=2.1=2\) (không phụ thuộc vào \(\alpha\))

\(\Rightarrow\left(đpcm\right)\)

\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

\(\Leftrightarrow B=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2\alpha+cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)

\(\Leftrightarrow B=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha\)

\(\Leftrightarrow B=\left(sin^2\alpha+cos^2\alpha\right)^3=1^3=1\) (không phụ thuộc vào \(\alpha\) ) \(\Rightarrow\left(đpcm\right)\)

a/A = sin2 + 2. sin.cos + cos2 + sin2 -2cos.sin + cos2= 2

Tớ không biết ghi anpha nên .. bucminh

2 tháng 1 2022

\(sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

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

\(=sin^4\alpha+2sin^2\alpha.cos^2\alpha+cos^2\alpha\)

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

8 tháng 10 2018

=(sin2α)3 + (cos2α)3 + 3sin2α - cos2α

= (sin2α + cos2α)(sin4α - sin2α.cos2α + cos4α) + 3sin2α - cos2α

= 1.(sin4α - sin2α.cos2α + cos4α) + 3sin2α - cos2α

= (1- cos2α) - (1- cos2α).cos2α + cos4α + 3(1- cos2α) - cos2α

[ có 1- cos2α là vì sin2α + cos2α = 1 => sin2α = 1- cos2α nên thay sin2α thành 1- cos2α ]

= 1 - 2cos2α + cos4α - cos2α + cos4α + cos4α + 3 - 3cos2α - cos2α

= 4 - 7cos2α + 3cos4α [rút vậy chắc gọn rồi ha =w=]