Rút gọn biểu thức:
a, cos2x - 4sin2\(\frac{x}{2}\)cos2\(\frac{x}{2}\)
b, \(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}\)
c, \(\frac{cosx+cos2x+cos3x+cos4x}{sinx+sin2x+sin3x+sin4x}\)
K
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R
Rút gọn
A= \(\frac{cosx-cos2x-cos3x+cos4x}{sinx-sin2x-sin3x+sin4x}\)
B= sinx(1+2cos2x+2cos4x+2cos6x)
1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2019
\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)
\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)
\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)
\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)
\(=sin7x\)
PN
1
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2019
\(A=\frac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\frac{sin2x}{cos2x}=tan2x\)
DM
2
2
6 tháng 5 2022
`A=[sin x + sin 2x + sin 3x]/[cos x + cos 2x + cos 3x]`
`A=[2sin2x.cosx+sin2x]/[2cos2x.cosx+cos2x]`
`A=[sin2x(2cosx+1)]/[cos2x(2cosx+1)]`
`A=tan 2x`
6 tháng 5 2022
\(A=\dfrac{sinx-sin2x+sin3x}{cosx-cos2x+cos3x}\)
\(ĐK\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\) \(A=\dfrac{sinx+sin3x-sin2x}{cosx+cos3x-cos2x}\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}=\dfrac{2sin2x.cosx-sin2x}{2cos2x.cosx-cos2x}\\=\dfrac{sin2x\left(2cosx-1\right)}{cos2x\left(2cosx-1\right)}\end{matrix}\right.\) \(\Rightarrow\) \(A=tan2x\)
21 tháng 7 2022
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
TH
9 tháng 8 2019
\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)
\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)
TP
1
TD
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2020
ĐKXĐ: ..
\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)
\(\Leftrightarrow tan2x=\sqrt{3}\)
\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)
29 tháng 9 2020
a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)
\(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=\frac{2sin2x}{sin2x}=2\)
\(\frac{cosx+cos3x+cos2x+cos4x}{sinx+sin3x+sin2x+sin4x}=\frac{2cosx.cos2x+2cosx.cos3x}{2sin2x.cosx+2sin3x.cosx}=\frac{2cosx\left(cos2x+cos3x\right)}{2cosx\left(sin2x+sin3x\right)}\)
\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)