Tính C = \(cos\frac{\Pi}{9}+cos\frac{2\Pi}{9}+....+cos\frac{8\Pi}{9}+cos\Pi\)
A . 0
B . 1
C . 2
D . -1
K
Khách
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
Ta có:
\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
\(B = \left( {\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9}} \right) + \cos \frac{{11\pi }}{9} = \left( {2\cos \frac{{\frac{\pi }{9} + \frac{{5\pi }}{9}}}{2}\cos \frac{{\frac{\pi }{9} - \frac{{5\pi }}{9}}}{2}} \right) + \cos \frac{{11\pi }}{9} = 2\cos \frac{\pi }{3}\cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9}\)
\( = \cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9} = 2\cos \frac{{\frac{{2\pi }}{9} + \frac{{11\pi }}{9}}}{2}\cos \frac{{\frac{{2\pi }}{9} - \frac{{11\pi }}{9}}}{2} = 2\cos \frac{{13\pi }}{{18}}\cos \frac{\pi }{2} = 0\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
c/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)
\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Bạn tự tìm x thuộc khoảng đã cho
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
b/
ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow tan^22x+1+tan^22x=7\)
\(\Leftrightarrow tan^22x=3\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)
Bạn tự tìm nghiệm thuộc khoảng đã cho nhé
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2020
\(sina.sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)\)
\(=-\frac{1}{2}sina\left[cos\frac{2\pi}{3}-cos2a\right]=-\frac{1}{2}sina\left(-\frac{1}{2}-cos2a\right)\)
\(=\frac{1}{4}sina+\frac{1}{2}sina.cos2a=\frac{1}{4}sina+\frac{1}{4}sin3a-\frac{1}{4}sina\)
\(=\frac{1}{4}sin3a\)
\(sin\frac{\pi}{9}sin\frac{2\pi}{9}sin\frac{4\pi}{9}=sin\frac{\pi}{9}sin\left(\frac{\pi}{3}-\frac{\pi}{9}\right)sin\left(\frac{\pi}{3}+\frac{\pi}{9}\right)=\frac{1}{4}sin\frac{\pi}{3}=\frac{\sqrt{3}}{8}\)
\(cosa.cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)=\frac{1}{2}cosa\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=\frac{1}{2}cosa\left(cos2a-\frac{1}{2}\right)=\frac{1}{2}cosa.cos2a-\frac{1}{4}cosa\)
\(=\frac{1}{4}cos3a+\frac{1}{4}cosa-\frac{1}{4}cosa=\frac{1}{4}cos3a\)
\(cos\frac{\pi}{18}cos\frac{5\pi}{18}cos\frac{7\pi}{18}=cos\frac{\pi}{18}.cos\left(\frac{\pi}{3}-\frac{\pi}{18}\right).cos\left(\frac{\pi}{3}+\frac{\pi}{18}\right)=\frac{1}{4}cos\frac{\pi}{6}=\frac{\sqrt{3}}{8}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2020
\(A=cos\left(32^0+28^0\right)=cos60^0=\frac{1}{2}\)
\(B=cos\left(220^0+170^0\right)=cos390^0=cos\left(30^0+360^0\right)=cos30^0=\frac{\sqrt{3}}{2}\)
\(C=sin\left(\frac{7\pi}{18}-\frac{5\pi}{9}\right)=sin\left(-\frac{\pi}{6}\right)=-sin\left(\frac{\pi}{6}\right)=-\frac{1}{2}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(A = \frac{{\sin \frac{\pi }{{15}}\cos \frac{\pi }{{10}} + \sin \frac{\pi }{{10}}\cos \frac{\pi }{{15}}}}{{\cos \frac{{2\pi }}{{15}}\cos \frac{\pi }{5} - \sin \frac{{2\pi }}{{15}}\sin \frac{\pi }{5}}} = \frac{{\sin \left( {\frac{\pi }{{15}} + \frac{\pi }{{10}}} \right)}}{{\cos \left( {\frac{{2\pi }}{{15}} + \frac{\pi }{5}} \right)}} = \frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{3}}} = 1\)
b) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)
\(C=cos\frac{\pi}{9}+cos\frac{8\pi}{9}+cos\frac{2\pi}{9}+cos\frac{7\pi}{9}+...+cos\frac{4\pi}{9}+cos\frac{5\pi}{9}+cos\pi\)
\(C=cos\frac{\pi}{9}+cos\left(\pi-\frac{\pi}{9}\right)+cos\frac{2\pi}{9}+cos\left(\pi-\frac{2\pi}{9}\right)+...+cos\frac{4\pi}{9}+cos\left(\pi-\frac{4\pi}{9}\right)-1\)
\(C=cos\frac{\pi}{9}-cos\frac{\pi}{9}+cos\frac{2\pi}{9}-cos\frac{2\pi}{9}+...+cos\frac{4\pi}{9}-cos\frac{4\pi}{9}-1\)
\(C=-1\)