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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}\)
a: \(A=\cos\alpha\cdot\sin^3\alpha+\cos^3\alpha\cdot\sin\alpha\)
\(=\dfrac{4}{5}\cdot\dfrac{27}{125}+\dfrac{64}{125}\cdot\dfrac{3}{5}\)
\(=\dfrac{4\cdot27+64\cdot3}{625}\)
\(=\dfrac{300}{625}=\dfrac{12}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
C = sin 4 α + cos 4 α = sin 4 α + cos 4 α + 2 sin 2 α . cos 2 α - 2 sin 2 α . cos 2 α
= sin 2 α + cos 2 α 2 - 2 sin 2 α . cos 2 α
= 1 - 2 sin 2 α . cos 2 α ( v ì sin 2 α + cos 2 α = 1 )
Vậy C = 1 - 2 sin 2 α . c o s 2
Đáp án cần chọn là: A
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)
=2
A = \(\left(sin^2a+cos^2a\right)^2=1^2=1\)
D = \(sin^2\left(sin^2B+cos^2B\right)+cos^2a=sin^2a+cos^2a=1\)