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dia fic
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Nguyễn Việt Lâm
21 tháng 4 2021 lúc 21:41

\(A=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}.cos\dfrac{5\pi}{11}.cos\left(\pi-\dfrac{4\pi}{11}\right)cos\left(\pi-\dfrac{2\pi}{11}\right)\)

\(=cos\dfrac{\pi}{11}.cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\left(-cos\dfrac{4\pi}{11}\right)\left(-cos\dfrac{2\pi}{11}\right)\)

\(=cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{5\pi}{11}\)

\(\Rightarrow2A.sin\dfrac{\pi}{11}=2sin\dfrac{\pi}{11}cos\dfrac{\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=sin\dfrac{2\pi}{11}cos\dfrac{2\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=\dfrac{1}{2}sin\dfrac{4\pi}{11}cos\dfrac{4\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{5\pi}{11}\)

\(=\dfrac{1}{4}sin\dfrac{8\pi}{11}.cos\dfrac{3\pi}{11}.cos\left(\pi-\dfrac{6\pi}{11}\right)\)

\(=-\dfrac{1}{4}sin\left(\pi-\dfrac{3\pi}{11}\right)cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{4}sin\dfrac{3\pi}{11}cos\dfrac{3\pi}{11}cos\dfrac{6\pi}{11}\)

\(=-\dfrac{1}{8}sin\dfrac{6\pi}{11}cos\dfrac{6\pi}{11}=-\dfrac{1}{16}sin\dfrac{12\pi}{11}=-\dfrac{1}{16}sin\left(\pi+\dfrac{\pi}{11}\right)\)

\(=\dfrac{1}{16}sin\dfrac{\pi}{11}\)

\(\Rightarrow A=\dfrac{1}{32}\)

Le van a
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Ngô Thành Chung
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Lê Thị Thục Hiền
27 tháng 5 2021 lúc 10:27

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 \(A=\dfrac{Y}{U}=4\)

Etermintrude💫
27 tháng 5 2021 lúc 10:10

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CHÚC BẠN HỌC TỐT NHÉok

Nguyễn Linh Chi
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xữ nữ của tôi
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camcon
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Nguyễn Lê Phước Thịnh
16 tháng 8 2023 lúc 22:22

\(=2\cdot cos\left(\dfrac{\dfrac{3}{5}pi+\dfrac{1}{5}pi}{2}\right)\cdot cos\left(\dfrac{\dfrac{3}{5}pi-\dfrac{1}{5}pi}{2}\right)\)

\(=2\cdot cos\left(\dfrac{2}{5}pi\right)\cdot cos\left(\dfrac{1}{5}pi\right)\)

James Pham
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Nguyễn Lê Phước Thịnh
30 tháng 7 2023 lúc 22:54

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

Trần MInh Hiển
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Lê Thị Thục Hiền
3 tháng 6 2021 lúc 20:37

\(cos\dfrac{\pi}{5}-cos\dfrac{2\pi}{5}\)

\(=-2.sin\dfrac{3\pi}{10}.sin\left(-\dfrac{\pi}{10}\right)\)

\(=2.sin\left(\dfrac{1}{2}-\dfrac{\pi}{5}\right).sin\dfrac{\pi}{10}\)

\(=2.sin\dfrac{\pi}{10}.cos\dfrac{\pi}{5}=\dfrac{sin\dfrac{\pi}{5}.cos\dfrac{\pi}{5}}{cos\dfrac{\pi}{10}}\)

\(=\dfrac{\dfrac{1}{2}sin\dfrac{2\pi}{5}}{cos\left(\dfrac{\pi}{2}-\dfrac{2\pi}{5}\right)}=\dfrac{\dfrac{1}{2}.sin\dfrac{2\pi}{5}}{sin\dfrac{2\pi}{5}}\)\(=\dfrac{1}{2}\)

=−2.sin3π10.sin(−π10)=−2.sin3π10.sin(−π10)

=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10=2.sinπ10.cosπ5=sinπ5.cosπ5cosπ10

=12

títtt
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Nguyễn Lê Phước Thịnh
22 tháng 8 2023 lúc 20:30

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)

Măm Măm
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