Câu 6.53 trang 205 SBT Đại số 10 Nâng cao

LG a

\(4\cos {15^0}\cos {21^0}\cos {24^0} - \cos {12^0} - \cos {18^0}\)

\(= \dfrac{{1 + \sqrt 3 }}{2}\);

Phương pháp giải:

Công thức: 

+) \(\cos a.\cos b = \frac{1}{2}(\cos (a + b) + \cos (a - b))\)

+) \(\cos a + \cos b = 2\cos \frac{{a + b}}{2}.\cos \frac{{a - b}}{2}\)

Lời giải chi tiết:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
2\cos {15^o}(2\cos {21^o}\cos {24^o}) - (\cos {12^o} + \cos {18^o})\\
= 2\cos {15^o}\left( {\cos \left( {{{21}^o} + {{24}^o}} \right) + \cos \left( {{{24}^o} - {{21}^o}} \right)} \right) - (2\cos \frac{{{{18}^o} + {{12}^o}}}{2}.\cos \frac{{{{18}^o} - {{12}^o}}}{2})
\end{array}\\
{ = 2\cos {{15}^o}\left( {\cos {{45}^o} + \cos {3^o}} \right) - 2\cos {{15}^o}\cos {3^o}}\\
{ = 2\cos {{15}^o}\cos {{45}^o} = \cos \left( {{{15}^o} + {{45}^o}} \right) + \cos \left( {{{45}^o} - {{15}^o}} \right)}\\
{ = \cos {{60}^o} + \cos {{30}^o} = \frac{1}{2} + \frac{{\sqrt 3 }}{2}.}
\end{array}\)

LG b

\(\tan {30^0} + \tan {40^0} + \tan {50^0} + \tan {60^0}\)

\(= \dfrac{{8\sqrt 3 }}{3}\cos {20^0}\);

Phương pháp giải:

Công thức: 

+) \(\tan a \pm \tan b = \frac{{\sin (a \pm b)}}{{\cos a.\cos b}}\)

+) \(\cos a.\cos b = \frac{1}{2}(\cos (a + b) + \cos (a - b))\)

+) \(\cos a + \cos b = 2\cos \frac{{a + b}}{2}.\cos \frac{{a - b}}{2}\)

Lời giải chi tiết:

\(\begin{array}{*{20}{l}}
{A = \left( {\tan {{30}^0} + \tan {{60}^0}} \right) + \left( {\tan {{40}^0} + \tan {{50}^0}} \right)}\\
\begin{array}{l}
= \frac{{\sin {{90}^0}}}{{\cos {{30}^0}\cos {{60}^0}}} + \frac{{\sin {{90}^0}}}{{\cos {{40}^0}\cos {{50}^0}}}\\
= \frac{1}{{\cos {{30}^0}\cos {{60}^0}}} + \frac{1}{{\cos {{40}^0}\cos {{50}^0}}}\\
= \frac{1}{{\frac{1}{2}\left( {\cos {{90}^0} + \cos {{30}^0}} \right)}} + \frac{1}{{\frac{1}{2}\left( {\cos {{90}^0} + \cos {{10}^0}} \right)}}\\
= 2.\left( {\frac{1}{{\cos {{30}^0}}} + \frac{1}{{\cos {{10}^0}}}} \right) = 2.\frac{{\cos {{30}^0} + \cos {{10}^0}}}{{\cos {{30}^0}.\cos {{10}^0}}}\\
= 2.\frac{{2.\cos {{20}^0}.\cos {{10}^0}}}{{\cos {{30}^0}.\cos {{10}^0}}} = 4.\frac{{\cos {{20}^0}}}{{\cos {{30}^0}}}\\
= 4.\frac{{\cos {{20}^0}}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{8\sqrt 3 }}{3}\cos {20^0}
\end{array}
\end{array}\)

LG c

\(\dfrac{1}{{\sin {{18}^0}}} - \dfrac{1}{{\sin {{54}^0}}} = 2;\)

Phương pháp giải:

+ \(\sin a - \sin b = 2\cos \frac{{a + b}}{2}\sin \frac{{a - b}}{2}\)

+ \(\sin a = \cos ({90^o} - a)\)

Lời giải chi tiết:

\(\begin{array}{l}\dfrac{1}{{\sin {{18}^0}}} - \dfrac{1}{{\sin {{54}^0}}} = \dfrac{{\sin {{54}^0} - \sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}}\\ = \dfrac{{2\cos {{36}^0}\sin {{18}^0}}}{{\sin {{18}^0}\sin {{54}^0}}} = \dfrac{{2\cos {{36}^0}}}{{\sin {{54}^0}}}\\ = \dfrac{{2\cos {{36}^0}}}{{\cos {{36}^0}}} = 2.\end{array}\)

LG d

\(\tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0} = 4\).

Phương pháp giải:

\(\tan {81^0} = \cot {90^0 - 81^0}= \cot {9^0}\)

\(\tan {81^0} = \cot {90^0 - 81^0}= \cot {9^0}\)

\(\cos {{36}^o}= \sin {90^0 - 36^0}= \sin {{54}^o}\)

\(\sin a - \sin b = 2.\cos \frac{{a + b}}{2}.\sin \frac{{a - b}}{2}\)

Lời giải chi tiết:

\(D= \tan {9^0} - \tan {27^0} - \tan {63^0} + \tan {81^0}\\ = \tan {9^0} + \tan {81^0} - \left( {\tan {{27}^0} + \tan {{63}^0}} \right)\)

\(\begin{array}{l}
\tan {9^o} + \tan {81^o} = \tan {9^o} + \cot {9^o}\\
= \frac{{\sin {9^o}}}{{\cos {9^o}}} + \frac{{\cos {9^o}}}{{\sin {9^o}}} = \frac{{{{\sin }^2}{9^o} + {{\cos }^2}{9^o}}}{{\sin {9^o}.\cos {9^o}}} = \frac{2}{{\sin {{18}^o}}}
\end{array}\)

Tương tự: \(\tan {{27}^0} + \tan {{63}^0}= \frac{2}{{\sin {{54}^o}}}\)

\(\begin{array}{l}
\Rightarrow D = 2\left( {\dfrac{1}{{\sin {{18}^o}}} - \dfrac{1}{{\sin {{54}^o}}}} \right) = 2.\dfrac{{\sin {{54}^o} - \sin {{18}^o}}}{{\sin {{54}^o}.\sin {{18}^o}}}\\
= 2.\dfrac{{2.\cos \frac{{{{54}^o} + {{18}^o}}}{2}.\sin \frac{{{{54}^o} - {{18}^o}}}{2}}}{{\sin {{54}^o}.\sin {{18}^o}}} = 4.\dfrac{{\cos {{36}^o}.\sin {{18}^o}}}{{\sin {{54}^o}.\sin {{18}^o}}} = 4
\end{array}\) 

(vì \(\cos {{36}^o}= \sin {90^0 - 36^0}= \sin {{54}^o}\))

Loigiaihay.com