Chứng minh rằng:
\(\frac{\sin^22x-4\sin^2x}{\sin^22x+4\sin^2x-4}=tan^4x\)
(Mấy bồ ơi, giúp với nè!)
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NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2019
\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)
\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)
\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 5 2020
\(\frac{sin^22x-4sin^2x}{sin^22x+4sin^2x-4}=\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4\left(sin^2x-1\right)}\)
\(=\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x.cos^2x-4cos^2x}=\frac{-4sin^4x}{4cos^2x\left(sin^2x-1\right)}=\frac{sin^4x}{cos^4x}=tan^4x\)
CT
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4sin^2x}+1=2tan^2x\)
\(\Leftrightarrow\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x\left(cos^2x+1\right)}+1=\frac{2sin^2x}{cos^2x}\)
\(\Leftrightarrow\frac{cos^2x}{cos^2x+1}=\frac{1-cos^2x}{cos^2x}\)
Đặt \(cos^2x=t\Rightarrow0< t< 1\)
\(\Rightarrow\frac{t}{t+1}=\frac{1-t}{t}\Leftrightarrow t^2=1-t^2\Leftrightarrow t^2=\frac{1}{2}\)
\(\Leftrightarrow t=\frac{\sqrt{2}}{2}\Leftrightarrow cos^2x=\frac{\sqrt{2}}{2}\)
18 tháng 8 2020
bạn ơi, cho mình hỏi là tại sao từ bước 2 xuống bước 3, tử sin22x-2 lại đổi thành 2-sin22x vậy ạ
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
Nhân cả tử và mẫu với -1 thôi bạn
\(=\frac{2-sin^22x}{4cos^2x\left(1-sin^2x\right)}=\frac{2-sin^2x}{4cos^2x.cos^2x}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 9 2020
b.
\(\Leftrightarrow2sin^2x+4sinx=3\left(1-sin^2x\right)\)
\(\Leftrightarrow5sin^2x+4sinx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{-2-\sqrt{19}}{5}\left(l\right)\\sinx=\frac{-2+\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2+\sqrt{19}}{5}\right)+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow sinx\left(sin^2x+3sinx+2\right)=0\)
\(\Leftrightarrow sinx\left(sinx+1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 9 2020
a.
\(1-cos^22x-\left(\frac{1-cos2x}{2}\right)=\frac{1}{2}\)
\(\Leftrightarrow2cos^22x-cos2x=0\)
\(\Leftrightarrow cos2x\left(2cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=\frac{1}{2}\\\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
a.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)
\(\Leftrightarrow1-sin^2x=0\)
\(\Leftrightarrow cos^2x=0\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
b.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
28 tháng 6 2021
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
HT
28 tháng 11 2019
Quên cách giải ptlg rồi nên lm câu 4 =.=
\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)
\(=\left(2\cos^2x-1\right)\cos x-2\sin^2x.\cos x\)
\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)
\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos x\left(2\cos^2x+\cos x-1\right)}{2\cos^2x-1+\cos x}=2\cos x\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
1.
\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)
\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)
\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)
\(\Leftrightarrow2cos^24x-cos4x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
3.
Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)
\(\Leftrightarrow1+4cos^22t=1-cos6t\)
\(\Leftrightarrow cos6t+4cos^22t=0\)
\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)
\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
\(\frac{sin^22x-4sin^2x}{sin^22x-4\left(1-sin^2x\right)}=\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x-4cos^2x}=\frac{sin^2x\left(cos^2x-1\right)}{cos^2x\left(sin^2x-1\right)}=\frac{-sin^4x}{-cos^4x}=tan^4x\)