CMR \(cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}=\dfrac{1}{2}\)
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27 tháng 5 2021
A\(=\dfrac{cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}}{cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{7}.cos\dfrac{5\pi}{7}}\)
Đặt tử là Y; mẫu là U
Có \(Y=\)\(cos\dfrac{5\pi}{7}.cos\dfrac{3\pi}{7}+\left(cos\dfrac{5\pi}{7}.cos\dfrac{\pi}{7}+cos\dfrac{3\pi}{7}.cos\dfrac{\pi}{7}\right)\)
\(=cos\left(\pi-\dfrac{2\pi}{7}\right).cos\left(\pi-\dfrac{4\pi}{7}\right)+cos\dfrac{\pi}{7}\left(cos\dfrac{5\pi}{7}+cos\dfrac{3\pi}{7}\right)\)
\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{\pi}{7}.2cos\dfrac{4\pi}{7}.cos\dfrac{\pi}{7}\)\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+2.cos^2\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}\)
\(=cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+\left(cos\dfrac{2\pi}{7}+1\right).cos\dfrac{4\pi}{7}\)\(=2.cos\dfrac{2\pi}{7}.cos\dfrac{4\pi}{7}+cos\dfrac{4\pi}{7}\)
\(=cos\dfrac{6\pi}{7}+cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}\)
\(\Rightarrow sin\dfrac{\pi}{7}.Y=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)
\(=\dfrac{1}{2}\left(-sin\dfrac{\pi}{7}+sin\dfrac{3\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{3\pi}{7}+sin\dfrac{5\pi}{7}\right)+\dfrac{1}{2}\left(-sin\dfrac{5\pi}{7}+sin\pi\right)\)
\(=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)\)\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\)
\(\Rightarrow Y=-\dfrac{1}{2}\)
Có \(sin\dfrac{\pi}{7}.U=sin\dfrac{\pi}{7}.cos\dfrac{\pi}{7}.cos\dfrac{3\pi}{5}.cos\dfrac{5\pi}{7}\)
\(=\dfrac{1}{2}.sin\dfrac{2\pi}{7}.cos\left(\pi-\dfrac{2\pi}{7}\right).cos\dfrac{3\pi}{5}\)
\(=-\dfrac{1}{4}.sin\dfrac{4\pi}{7}.cos\left(\pi-\dfrac{4\pi}{5}\right)\)
\(=\dfrac{1}{8}.sin\dfrac{8\pi}{7}\)\(=\dfrac{1}{8}.sin\left(\pi+\dfrac{\pi}{7}\right)=-\dfrac{1}{8}.sin\dfrac{\pi}{7}\)
\(\Rightarrow U=-\dfrac{1}{8}\)
Vậy \(A=\dfrac{Y}{U}=4\)
30 tháng 7 2023
d: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
a: =>sin 4x=cos(x+pi/6)
=>sin 4x=sin(pi/2-x-pi/6)
=>sin 4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi
=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi
c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)
\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)
\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)
\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)
\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)
\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)
\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)
\(\Rightarrow A=\dfrac{1}{32}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
22 tháng 8 2023
a: cos3x=8
mà -1<=cos3x<=1
nên \(x\in\varnothing\)
b; \(-2\cdot cosx+\sqrt{3}=0\)
=>\(-2\cdot cosx=-\sqrt{3}\)
=>\(cosx=\dfrac{\sqrt{3}}{2}\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: cos(3x-pi/6)=0
=>3x-pi/6=pi/2+k2pi
=>3x=2/3pi+k2pi
=>x=2/9pi+k2pi/3
d: cos(x+2/3pi)=cos(pi/5)
=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi
=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi
e: cos^2(3x)=4
=>cos3x=2(loại) hoặc cos3x=-2(loại)
\(cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}\)
\(=-2.sin\dfrac{3\pi}{10}.sin\left(-\dfrac{\pi}{10}\right)\)
\(=2.sin\left(\dfrac{1}{2}-\dfrac{\pi}{5}\right).sin\dfrac{\pi}{10}\)
\(=2.sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}=\dfrac{sin\dfrac{\pi}{5}.cos\dfrac{\pi}{5}}{cos\dfrac{\pi}{10}}\)
\(=\dfrac{\dfrac{1}{2}sin\dfrac{2\pi}{5}}{cos\left(\dfrac{\pi}{2}-\dfrac{2\pi}{5}\right)}=\dfrac{\dfrac{1}{2}.sin\dfrac{2\pi}{5}}{sin\dfrac{2\pi}{5}}\)\(=\dfrac{1}{2}\)
cosπ5−cos2π5cosπ5−cos2π5
=−2.sin3π10.sin(−π10)=−2.sin3π10.sin(−π10)
=2.sin(12−π5).sinπ10=2.sin(12−π5).sinπ10
=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10
=12sin2π5cos(π2−2π5)=12.sin2π5sin2π5=12sin2π5cos(π2−2π5)=12.sin2π5sin2π5=12