MN K BT?

\(\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}\)

a: \(A=\cos\alpha\cdot\sin^3\alpha+\cos^3\alpha\cdot\sin\alpha\)

\(=\dfrac{4}{5}\cdot\dfrac{27}{125}+\dfrac{64}{125}\cdot\dfrac{3}{5}\)

\(=\dfrac{4\cdot27+64\cdot3}{625}\)

\(=\dfrac{300}{625}=\dfrac{12}{25}\)

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Trở về chém ổng đứt làm ba

thấy hay thì

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

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

\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)

Vậy...

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

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

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

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

Vậy...

3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)

\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)

\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)

Vậy...

4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)

\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)

\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)

Vậy...

5, Xem lại đề

6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)

\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)

Vậy...

a) \(A=2\left(sin^6\alpha+cos^6\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)\)\(-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=-\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)^2=-1\) (Không phụ thuộc vào \(\alpha\)).

b) \(B=4\left(sin^4\alpha+cos^4\alpha\right)-cos4\alpha\)
\(=4\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)-8sin^2\alpha cos^2\alpha\)\(-\left(1-2sin^22\alpha\right)\)
\(=4.\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^22\alpha-1+2sin^22\alpha\)
\(=4-1=3\).

c) \(8\left(cos^8\alpha-sin^8\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^4\alpha-sin^4\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^2\alpha-sin^2\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-8cos2\alpha+cos2\alpha-cos6\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4-1\right)+sin4\alpha sin2\alpha\)
\(=8cos2\alpha\left[\left(sin^4\alpha-sin^2\alpha\right)+\left(cos^4\alpha-cos^2\alpha\right)\right]+\)\(sin4\alpha sin2\alpha\)
\(=8cos2\alpha.\left[sin^2\alpha\left(sin^2\alpha-1\right)+cos^2\alpha\left(cos^2\alpha-1\right)\right]\)\(+sin4\alpha sin2\alpha\)
\(=8.cos2\alpha.\left(-2sin^2\alpha cos^2\alpha\right)+2sin2\alpha cos2\alpha sin2\alpha\)
\(=-2cos2\alpha.\left(sin2\alpha\right)^2+2cos2\alpha.\left(sin2\alpha\right)^2\)
\(=sin^22\alpha\left(-cos2\alpha+cos2\alpha\right)=sin^22\alpha.0=0\) (không phụ thuộc vào \(\alpha\)).

a)
\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1+2sin\alpha cos\alpha+1-2sin\alpha cos\alpha=2\) (không phụ thuộc vào \(\alpha\)).
b)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).