Tính: P=\(\cos\frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}...\cos\frac{x}{2^n}\).
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NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
1.
\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}\left(cos^4\frac{x}{2}-sin^4\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx.cosx=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow\frac{1}{2}sin2x=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sin2x=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
3.
ĐKXĐ: ...
\(\frac{1}{cosx}+\frac{1}{2sinx.cosx}=\frac{1}{2sinx.cosx.cos2x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow cos2x-sinx=0\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow2sin^2x+sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
7 tháng 4 2016
Ta có \(A=\frac{3}{2}+\frac{1}{2}\left[\cos2x+\cos\left(\frac{2\pi}{3}+2x\right)+\cos\left(\frac{4\pi}{3}+2x\right)\right]\)
\(=\frac{3}{2}+\frac{1}{2}\left[\cos2x+2\cos\left(\pi+2x\right).\cos\left(-\frac{\pi}{3}\right)\right]=\frac{3}{2}+\frac{1}{2}\left[\cos2x+\cos2x\right]=\frac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)
\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)
\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)
\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)
\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 2 2020
\(\Leftrightarrow4cosx\left(cos2x+cos\frac{4\pi}{3}\right)\left(cos6x+cos\frac{4\pi}{3}\right)=sin9x\)
\(\Leftrightarrow cosx\left(2cos2x-1\right)\left(2cos6x-1\right)=sin9x\)
\(\Leftrightarrow\left(2cos2x.cosx-cosx\right)\left(2cos6x-1\right)=sin9x\)
\(\Leftrightarrow\left(cos3x+cosx-cosx\right)\left(2cos6x-1\right)=sin9x\)
\(\Leftrightarrow cos3x\left(2cos6x-1\right)=sin9x\)
\(\Leftrightarrow2cos6x.cos3x-cos3x=sin9x\)
\(\Leftrightarrow cos9x+cos3x-cos3x=sin9x\)
\(\Leftrightarrow cos9x=sin9x\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 2 2020
\(cos\frac{4\pi}{3}=-\frac{1}{2}\), sau đó số 4 bên ngoài bạn biến thành 2 số 2 nhân vào 2 cái ngoặc là được thôi
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
7.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow8cosx=\frac{\sqrt{3}cosx+sinx}{sinx.cosx}\)
\(\Leftrightarrow8cosx.sinx.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow4sin2x.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)
\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\)
\(\Leftrightarrow sin\left(-3x\right)=sin\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x-\frac{\pi}{3}+k2\pi\\-3x=\frac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
5.
\(sin\left(2x+\frac{\pi}{2}+2\pi\right)-2cos\left(x+\frac{\pi}{2}-4\pi\right)=1+2sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)-2cos\left(x+\frac{\pi}{2}\right)=1+2sinx\)
\(\Leftrightarrow cos2x+2sinx=1+2sinx\)
\(\Leftrightarrow cos2x=1\)
\(\Rightarrow x=k\pi\)
6.
\(sin^22x-cos^28x=sin\left(10x+\frac{\pi}{2}+8\pi\right)\)
\(\Leftrightarrow\frac{1-cos4x}{2}-\frac{1+cos16x}{2}=sin\left(10x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow-\left(cos4x+cos16x\right)=2cos10x\)
\(\Leftrightarrow-2cos10x.cos6x=2cos10x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}10x=\frac{\pi}{2}+k\pi\\6x=\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\end{matrix}\right.\)
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.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 8 2020
\(\Leftrightarrow1-2sin^2\left(\frac{x}{2}+\frac{\pi}{6}\right)+1-2sin^2\frac{x}{2}=\frac{3}{2}-4sin\left(\frac{x}{2}\right)sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[sin\left(\frac{x}{2}\right)-sin\left(\frac{x}{2}+\frac{\pi}{6}\right)\right]^2=\frac{1}{4}\)
\(\Leftrightarrow4cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right).sin^2\left(\frac{\pi}{12}\right)=\frac{1}{4}\)
\(\Leftrightarrow cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right)=\frac{1}{16sin^2\left(\frac{\pi}{12}\right)}=\frac{2+\sqrt{3}}{4}\)
\(\Leftrightarrow1+cos\left(x+\frac{\pi}{6}\right)=\frac{2+\sqrt{3}}{2}\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)