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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 \(A=\dfrac{Y}{U}=4\)
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 \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2021
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2021
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
5 tháng 7 2021
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
\(cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}\)
\(=-2.sin\dfrac{3\pi}{10}.sin\left(-\dfrac{\pi}{10}\right)\)
\(=2.sin\left(\dfrac{1}{2}-\dfrac{\pi}{5}\right).sin\dfrac{\pi}{10}\)
\(=2.sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}=\dfrac{sin\dfrac{\pi}{5}.cos\dfrac{\pi}{5}}{cos\dfrac{\pi}{10}}\)
\(=\dfrac{\dfrac{1}{2}sin\dfrac{2\pi}{5}}{cos\left(\dfrac{\pi}{2}-\dfrac{2\pi}{5}\right)}=\dfrac{\dfrac{1}{2}.sin\dfrac{2\pi}{5}}{sin\dfrac{2\pi}{5}}\)\(=\dfrac{1}{2}\)
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