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1 tháng 8 2023
a: tan x(cot^2x-1)
\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)
=cotx-tanx/cotx=cotx(1-tan^2x)
b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)
c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)
\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)
=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)
=-cos^2x*cos^2x=-cos^4x
=>ĐPCM
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)-1\)
\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=cos\left(\dfrac{\pi}{2}-x\right)\)
\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sin\dfrac{x}{2}-cos\dfrac{x}{2}.sinx=1\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}.cos^2\dfrac{x}{2}=1\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\left(1-sin^2\dfrac{x}{2}\right)=1\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow\left(sin\dfrac{x}{2}-1\right)\left(2sin^2\dfrac{x}{2}+2sin\dfrac{x}{2}+1\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2020
\(sina+sinb+sinc+3=0\)
\(\Leftrightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)=0\)
Do \(\left\{{}\begin{matrix}sina\ge-1\\sinb\ge-1\\sinc\ge-1\end{matrix}\right.\) ;\(\forall a;b;c\)
\(\Rightarrow\left(sina+1\right)+\left(sinb+1\right)+\left(sinc+1\right)\ge0\)
Dấu "=" xảy ra khi và chỉ khi \(sina=sinb=sinc=-1\)
\(\Rightarrow cosa=cosb=cosc=0\Rightarrow cosa+cosb+cosc+10=10\)
b/ \(sinx=1-sin^2x\Rightarrow sinx=cos^2x\)
\(\Rightarrow sin^2x=cos^4x\Rightarrow1-cos^2x=cos^4x\)
\(\Rightarrow cos^4x+cos^2x=1\Rightarrow\left(cos^4x+cos^2x\right)^2=1\)
\(\Rightarrow cos^8x+2cos^6x+cos^4x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2020
\(1+sin5x=2cos^2\frac{x}{2}\)
\(\Leftrightarrow sin5x=2cos^2\frac{x}{2}-1\)
\(\Leftrightarrow sin5x=cosx\)
\(\Leftrightarrow sin5x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{2}-x+k2\pi\\5x=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2020
Câu đầu đơn giản là ko dịch được \(cos^22\times x/2\) nghĩa là gì :)
\(sin5x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow sin5x=-\left(cos^2x-sin^2x\right)\)
\(\Leftrightarrow sin5x=-cos2x\)
\(\Leftrightarrow sin5x=sin\left(2x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2x-\frac{\pi}{2}+k2\pi\\5x=\frac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
CM
4 tháng 3 2018
a) √2 cos(x - π/4)
= √2.(cosx.cos π/4 + sinx.sin π/4)
= √2.(√2/2.cosx + √2/2.sinx)
= √2.√2/2.cosx + √2.√2/2.sinx
= cosx + sinx (đpcm)
b) √2.sin(x - π/4)
= √2.(sinx.cos π/4 - sin π/4.cosx )
= √2.(√2/2.sinx - √2/2.cosx )
= √2.√2/2.sinx - √2.√2/2.cosx
= sinx – cosx (đpcm).
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 5 2020
3.
\(f\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{\pi}{3}\right)\Rightarrow f'\left(x+\frac{\pi}{3}\right)=-sin\left(x+\frac{\pi}{3}\right)\)
\(f'\left(x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{6}\right)\)
\(f'\left(0\right)=-sin\left(0\right)=0\)
\(2f'\left(x+\frac{\pi}{3}\right).f'\left(x-\frac{\pi}{6}\right)=2sin\left(x+\frac{\pi}{3}\right)sin\left(x-\frac{\pi}{6}\right)\)
\(=cos\left(\frac{\pi}{2}\right)-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)=0-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(\Rightarrow2f'\left(x+\frac{\pi}{3}\right)f'\left(x-\frac{\pi}{6}\right)=f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)\) (đpcm)
4.
\(y=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
\(=3\left(sin^2x+cos^2x\right)^2-6sin^2x.cos^2x-2\left(sin^2x+cos^2x\right)^3+6sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=3-2=1\)
\(\Rightarrow y'=0\) ; \(\forall x\)
5.
\(y=\left(\frac{sinx}{1+cosx}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{1-cos^2x}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{sin^2x}\right)^3=\left(\frac{1-cosx}{sinx}\right)^3\)
\(y'=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{sin^2x-cosx\left(1-cosx\right)}{sin^2x}\right)=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{1-cosx}{sin^2x}\right)=\frac{3\left(1-cosx\right)^3}{sin^4x}\)
\(\Rightarrow y'.sinx-3y=\frac{3\left(1-cosx\right)^3}{sin^3x}-3\left(\frac{1-cosx}{sinx}\right)^3=0\) (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 9 2020
e/
\(2cos^2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow1+cos2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow2cos^32x+cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x-1\right)\left(2cos^22x+3cos2x+2\right)=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
\(sin^4\left(x\right)+cos^4\left(x\right)+2sin^2\left(x\right)cos^2\left(x\right)=\left[sin^2\left(x\right)+cos^2\left(x\right)\right]^2=1^2=1\\ \Rightarrow sin^4\left(x\right)+cos^4\left(x\right)=1-2sin^2\left(x\right)cos^2\left(x\right)\left(đpcm\right)\)
sin^4x+cos^4x
=(sin^2x+cos^2x)^2-2*sin^2x*cos^2x
=1-2*cos^2x*sin^2x