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Lâm Ánh Yên
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Nguyễn Việt Lâm
6 tháng 8 2021 lúc 17:28

\(A=cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\left(-cos\left(\pi-\dfrac{5\pi}{7}\right)\right)=-cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(\Rightarrow A.sin\left(\dfrac{\pi}{7}\right)=-sin\left(\dfrac{\pi}{7}\right).cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=-\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)=-\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=-\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)=\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)

\(\Rightarrow A=\dfrac{1}{8}\)

\(B=\dfrac{\sqrt{3}}{2}.cos48^0.cos24^0.cos12^0\)

\(\Rightarrow B.sin12^0=\dfrac{\sqrt{3}}{2}sin12^0.cos12^0cos24^0.cos48^0\)

\(=\dfrac{\sqrt{3}}{4}sin24^0cos24^0cos48^0=\dfrac{\sqrt{3}}{8}sin48^0.cos48^0\)

\(=\dfrac{\sqrt{3}}{16}sin96^0=\dfrac{\sqrt{3}}{16}cos6^0\)

\(\Rightarrow2B.sin6^0.cos6^0=\dfrac{\sqrt{3}}{16}cos6^0\Rightarrow B=\dfrac{\sqrt{3}}{32.sin6^0}\)

Biểu thức này ko thể rút gọn tiếp được

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Kimian Hajan Ruventaren
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Nguyễn Việt Lâm
21 tháng 4 2021 lúc 16:43

a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

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Ngô Thành Chung
21 tháng 4 2021 lúc 18:22

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)

 

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Trần Công Thanh Tài
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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 \(A=\dfrac{Y}{U}=4\)

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Etermintrude💫
27 tháng 5 2021 lúc 10:10

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CHÚC BẠN HỌC TỐT NHÉok

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Trần Công Thanh Tài
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Nguyễn Lê Phước Thịnh
30 tháng 7 2023 lúc 20:38

\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)

\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)

\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)

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camcon
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Nguyễn Lê Phước Thịnh
16 tháng 8 2023 lúc 22:46

\(cos^2\left(\dfrac{pi}{7}\right)+cos^2\left(\dfrac{2pi}{7}\right)+cos^2\left(\dfrac{3pi}{7}\right)\)

\(=1-2\cdot cos\left(\dfrac{pi}{7}\right)\cdot cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{3pi}{7}\right)\)

\(=1-2\cdot\dfrac{1}{2}\left[cos\left(\dfrac{pi}{7}+\dfrac{3pi}{7}\right)+cos\left(\dfrac{3pi}{7}-\dfrac{pi}{7}\right)\right]\cdot cos\left(\dfrac{2pi}{7}\right)\)

\(=1-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{4pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{2pi}{7}\right)\)

\(=1-cos^2\left(\dfrac{2pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot cos\left(\dfrac{4pi}{7}\right)\)

\(=sin^2\left(\dfrac{2pi}{7}\right)-cos\left(\dfrac{2pi}{7}\right)\cdot\left[2\cdot cos^2\left(\dfrac{2pi}{7}\right)-1\right]\)

\(=sin^2\left(\dfrac{2pi}{7}\right)-2\cdot cos^3\left(\dfrac{2pi}{7}\right)+cos\left(\dfrac{2pi}{7}\right)\)

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Hobiee
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Le van a
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Kimian Hajan Ruventaren
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Lê Thùy Linh
2 tháng 5 2021 lúc 20:59

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