cho x là 1 góc nhọn , Rút gọn : sin^6x + cos^6x + 3sin^2x.cosx
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11 tháng 9 2017
A= \(\left(\sin^2a\right)^3+\left(cos^2a\right)^3+3sin^2acos^2a\)
=\(\left(sin^2a+cos^2a\right)\left(sin^4a-cos^2asin^2a+cos^4a\right)+3sin^2acos^2a\)
\(sin^4a+2sin^2acos^2a+cos^4a=\left(sin^2+cos^2\right)^2=1^2=1\)
NC
1
1 tháng 8 2021
Ta có: \(A=\sin^6x+3\cdot\sin^4x\cdot\cos^2x+3\cdot\sin^2x\cdot\cos^4x+\cos^6x\)
\(=\left(\sin^2x+\cos^2x\right)^3\)
=1
HH
0
5 tháng 7 2021
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...
7 tháng 9 2016
\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)
\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)
\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)
TH
1
TN
3 tháng 6 2017
= (sin^2x + cos^2x)^2 - 3sin^4x.cos^2x - 3sin^2x.cos^4x
= 1 - 3/4sin^2 (2x).sin^2x - 3/4sin^2(2x).cos^2x
= 1 - 3/4sin^2(2x)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
a.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)
\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
b.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
TN
1
MT
21 tháng 10 2015
\(sin^6x+cos^6x+3sin^2x.cos^2x=\left(sin^2x\right)^3+\left(cos^2x\right)^3+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)\left[\left(sin^2x\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2\right]+3sin^2x.cos^2x\)
\(=1.\left[\left(sin^2\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2\right]+3sin^2x.cos^2x\)
\(=\left(sin^2x\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2=1^2=1\)
dungf hddt thucws chosin^6x + cos ^6x