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NV
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
12 tháng 7 2020

\(A=cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7}=cos\frac{\pi}{7}cos\frac{4\pi}{7}cos\frac{2\pi}{7}\)

\(\Rightarrow A.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}cos\frac{4\pi}{7}\)

\(=\frac{1}{2}sin\frac{2\pi}{7}cos\frac{2\pi}{7}cos\frac{4\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}cos\frac{4\pi}{7}\)

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

\(\Rightarrow A=-\frac{1}{8}\)

\(B=sin6.cos48.cos24.cos12\)

\(B.cos6=sin6.cos6.cos12.cos24.cos48\)

\(=\frac{1}{2}sin12.cos12.cos24.cos48=\frac{1}{4}sin24.cos24.cos48\)

\(=\frac{1}{8}sin48.cos48=\frac{1}{16}sin96\)

\(=\frac{1}{16}sin\left(90+6\right)=\frac{1}{16}cos6\Rightarrow B=\frac{1}{16}\)

HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023

\(A = \cos {75^0}\cos {15^0} = \frac{1}{2}\left[ {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right] \\= \frac{1}{2}.\cos {60^0}.\cos {90^0} = 0\)

\(B = \sin \frac{{5\pi }}{{12}}\cos \frac{{7\pi }}{{12}} = \frac{1}{2}\left[ {\sin \left( {\frac{{5\pi }}{{12}} - \frac{{7\pi }}{{12}}} \right) + \sin \left( {\frac{{5\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right] \\= \frac{1}{2}\sin \left( { - \frac{{2\pi }}{{12}}} \right).\sin \left( {\frac{{12\pi }}{{12}}} \right) =  - \frac{1}{2}\sin \frac{\pi }{6}\sin \pi  = 0\)

HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023

\(B = \left( {\cos \frac{\pi }{9} + \cos \frac{{5\pi }}{9}} \right) + \cos \frac{{11\pi }}{9} = \left( {2\cos \frac{{\frac{\pi }{9} + \frac{{5\pi }}{9}}}{2}\cos \frac{{\frac{\pi }{9} - \frac{{5\pi }}{9}}}{2}} \right) + \cos \frac{{11\pi }}{9} = 2\cos \frac{\pi }{3}\cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9}\)

\( = \cos \frac{{2\pi }}{9} + \cos \frac{{11\pi }}{9} = 2\cos \frac{{\frac{{2\pi }}{9} + \frac{{11\pi }}{9}}}{2}\cos \frac{{\frac{{2\pi }}{9} - \frac{{11\pi }}{9}}}{2} = 2\cos \frac{{13\pi }}{{18}}\cos \frac{\pi }{2} = 0\)

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) \(B = \sin \frac{\pi }{{32}}\cos \frac{\pi }{{32}}\cos \frac{\pi }{{16}}\cos \frac{\pi }{8} = \frac{1}{2}\sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{8}.\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{1}{8}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{{16}}\;.\)

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
15 tháng 5 2020

\(sin\left(\frac{\pi}{7}\right)H=sin\left(\frac{\pi}{7}\right)cos\left(\frac{2\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{4\pi}{7}\right)+sin\left(\frac{\pi}{7}\right)cos\left(\frac{6\pi}{7}\right)\)

\(=\frac{1}{2}\left[sin\left(\frac{3\pi}{7}\right)-sin\left(\frac{\pi}{7}\right)+sin\left(\frac{5\pi}{7}\right)-sin\left(\frac{3\pi}{7}\right)+sin\pi-sin\left(\frac{5\pi}{7}\right)\right]\)

\(=-\frac{1}{2}sin\left(\frac{\pi}{7}\right)\)

\(\Rightarrow H=-\frac{1}{2}\)

\(sinA+sinB+sinC=2sin\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)+2sin\left(\frac{C}{2}\right)cos\left(\frac{C}{2}\right)\)

\(=2cos\frac{C}{2}cos\left(\frac{A-B}{2}\right)+2cos\left(\frac{A+B}{2}\right)cos\frac{C}{2}\)

\(=2cos\frac{C}{2}\left[cos\left(\frac{A-B}{2}\right)+cos\left(\frac{A+B}{2}\right)\right]\)

\(=4cos\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}\)

HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023

Ta có:

\(D = \frac{{\sin \frac{{7\pi }}{9} + \sin \frac{\pi }{9}}}{{\cos \frac{{7\pi }}{9} - \cos \frac{\pi }{9}}} = \frac{{2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\cos \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}}{{ - 2.\sin \left( {\frac{{\frac{{7\pi }}{9} + \frac{\pi }{9}}}{2}} \right).\sin \left( {\frac{{\frac{{7\pi }}{9} - \frac{\pi }{9}}}{2}} \right)}} = -\cot \frac{\pi }{3} = -\frac{{\sqrt 3 }}{3}\)