tính cos\(\frac{\pi}{8}\) và M=\(cos\frac{\pi}{7}\) -cos\(\frac{2\pi}{7}\) +cos\(\frac{3\pi}{7}\)
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
Ta có:
\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)
LL
0
24 tháng 8 2023
\(cos\dfrac{7\pi}{12}+cos\dfrac{\pi}{12}\\ =2.cos\dfrac{\dfrac{7\pi}{12}+\dfrac{\pi}{12}}{2}\\ =2.cos\dfrac{\pi}{3}.cos\dfrac{\pi}{4}\\ =2.\dfrac{1}{2}.\dfrac{\sqrt{2}}{2}\\ =\dfrac{\sqrt{2}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2019
\(cos\left(2\pi+\frac{\pi}{16}\right).sin\frac{5\pi}{16}.cos\frac{5\pi}{16}.cos\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\)
\(=\frac{1}{4}.2cos\frac{\pi}{16}.sin\frac{\pi}{16}.2sin\frac{5\pi}{16}.cos\frac{5\pi}{16}\)
\(=\frac{1}{4}sin\frac{2\pi}{16}.sin\frac{10\pi}{16}=\frac{1}{4}sin\frac{\pi}{8}.sin\frac{5\pi}{8}\)
\(=\frac{1}{4}sin\frac{\pi}{8}.sin\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\)
\(=\frac{1}{4}sin\frac{\pi}{8}.cos\frac{\pi}{8}=\frac{1}{8}sin\frac{2\pi}{8}\)
\(=\frac{1}{8}sin\frac{\pi}{4}=\frac{\sqrt{2}}{16}\)
Đề sai hoặc bạn gõ thiếu số 1 ở dưới mẫu
QT
Quoc Tran Anh Le
Giáo viên
21 tháng 9 2023
Ta có:
\(\begin{array}{l}\sin \frac{\pi }{{24}}\cos \frac{{5\pi }}{{24}} = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{{24}} + \frac{{5\pi }}{{24}}} \right) + \sin \left( {\frac{\pi }{{24}} - \frac{{5\pi }}{{24}}} \right)} \right]\\ = \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{4}} \right) + \sin \left( { - \frac{\pi }{6}} \right)} \right]\\ = \frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} - \frac{1}{2}} \right] = \frac{{\sqrt 2 - 1}}{4}\end{array}\)
Ta có:
\(\begin{array}{l}\sin \frac{{7\pi }}{8}\sin \frac{{5\pi }}{8} = \frac{1}{2}\left[ {\cos \left( {\frac{{7\pi }}{8} - \frac{{5\pi }}{8}} \right) - \cos \left( {\frac{{7\pi }}{8} + \frac{{5\pi }}{8}} \right)} \right]\\ = \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{4}} \right) - \cos \left( {\frac{{3\pi }}{2}} \right)} \right]\\ = \frac{1}{2}.\left( {\frac{{\sqrt 2 }}{2} + 0} \right) = \frac{{\sqrt 2 }}{4}\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2020
- Xét \(sin\frac{x}{5}=0\Rightarrow C=...\)
- Với \(sin\frac{x}{5}\ne0\)
\(C.sin\frac{x}{5}=sin\frac{x}{5}.cos\frac{x}{5}.cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{2}sin\frac{2x}{5}cos\frac{2x}{5}cos\frac{4x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{4}sin\frac{4x}{5}cos\frac{4x}{5}cos\frac{8x}{5}=\frac{1}{8}sin\frac{8x}{5}cos\frac{8x}{5}\)
\(=\frac{1}{16}sin\frac{16x}{5}\Rightarrow C=\frac{sin\frac{16x}{5}}{16.sin\frac{x}{5}}\)
\(D=sin\frac{x}{7}+sin\frac{5x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}cos\frac{2x}{7}+2sin\frac{3x}{7}\)
\(=2sin\frac{3x}{7}\left(cos\frac{2x}{7}+1\right)=4cos^2\frac{x}{7}.sin\frac{3x}{7}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2020
\(A=cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7}=cos\frac{\pi}{7}cos\frac{4\pi}{7}cos\frac{2\pi}{7}\)
\(\Rightarrow A.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{2}sin\frac{2\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}cos\frac{4\pi}{7}\)
\(=\frac{1}{8}sin\frac{8\pi}{7}=\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=-\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=-\frac{1}{8}\)
\(B=sin6.cos48.cos24.cos12\)
\(B.cos6=sin6.cos6.cos12.cos24.cos48\)
\(=\frac{1}{2}sin12.cos12.cos24.cos48=\frac{1}{4}sin24.cos24.cos48\)
\(=\frac{1}{8}sin48.cos48=\frac{1}{16}sin96\)
\(=\frac{1}{16}sin\left(90+6\right)=\frac{1}{16}cos6\Rightarrow B=\frac{1}{16}\)
\(cos\frac{\pi}{4}=2cos^2\frac{\pi}{8}-1\Rightarrow cos^2\frac{\pi}{8}=\frac{cos\frac{\pi}{4}+1}{2}\)
\(\Rightarrow cos^2\frac{\pi}{8}=\frac{2+\sqrt{2}}{4}\Rightarrow cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\) (do \(0< \frac{\pi}{8}< \frac{\pi}{2}\) nên \(cos\frac{\pi}{8}>0\))
\(M=cos\frac{\pi}{7}-cos\frac{2\pi}{7}+cos\frac{3\pi}{7}\)
\(\Rightarrow2M.sin\frac{\pi}{7}=2sin\frac{\pi}{7}cos\frac{\pi}{7}-2sin\frac{\pi}{7}cos\frac{2\pi}{7}+2sin\frac{\pi}{7}cos\frac{3\pi}{7}\)
\(=sin\frac{2\pi}{7}-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\frac{4\pi}{7}-sin\frac{2\pi}{7}\)
\(=-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\left(\pi-\frac{3\pi}{7}\right)\)
\(=-sin\frac{3\pi}{7}+sin\frac{\pi}{7}+sin\frac{3\pi}{7}=sin\frac{\pi}{7}\)
\(\Rightarrow M=\frac{sin\frac{\pi}{7}}{2sin\frac{\pi}{7}}=\frac{1}{2}\)