Bài 23 trang 224 SGK Đại số 10 Nâng cao

LG a

\({\sin ^2}({\pi  \over 8} + \alpha ) - {\sin ^2}({\pi  \over 8} - \alpha ) \) \(= {{\sqrt 2 } \over {2 }}\sin 2\alpha\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& {\sin ^2}({\pi \over 8} + \alpha ) - {\sin ^2}({\pi \over 8} - \alpha ) \cr&= {\rm{[}}\sin ({\pi \over 8} + \alpha ) + \sin ({\pi \over 8} - \alpha ){\rm{]}}.\cr&\;\;\;\;\;{\rm{[}}\sin ({\pi \over 8} + \alpha ) - \sin ({\pi \over 8} - \alpha ){\rm{]}} \cr 
& {\rm{ = (2sin}}{\pi \over 8}\cos \alpha )(2\cos {\pi \over 8}\sin \alpha ) \cr& = 2\sin \frac{\pi }{8}\cos \frac{\pi }{8}.2\sin \alpha \cos \alpha \cr &= \sin {\pi \over 4}\sin 2\alpha = {{\sqrt 2 } \over {2 }} \sin 2\alpha \cr} \) 

LG b

\({\cos ^2}\alpha  + {\cos ^2}(\alpha  - {\pi  \over 3}) + {\cos ^2}({{2\pi } \over 3} - \alpha ) \) \(= {3 \over 2}\)

Lời giải chi tiết:

\[\begin{array}{l}
{\cos ^2}\alpha + {\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right) + {\cos ^2}\left( {\frac{{2\pi }}{3} - \alpha } \right)\\
= {\cos ^2}\alpha + {\left( {\cos \alpha \cos \frac{\pi }{3} + \sin \alpha \sin \frac{\pi }{3}} \right)^2}\\
+ {\left( {\cos \frac{{2\pi }}{3}\cos \alpha + \sin \frac{{2\pi }}{3}\sin \alpha } \right)^2}\\
= {\cos ^2}\alpha + {\left( {\frac{1}{2}\cos \alpha + \frac{{\sqrt 3 }}{2}\sin \alpha } \right)^2}\\
+ {\left( { - \frac{1}{2}\cos \alpha + \frac{{\sqrt 3 }}{2}\sin \alpha } \right)^2}\\
= {\cos ^2}\alpha + \frac{1}{4}{\cos ^2}\alpha + \frac{3}{4}{\sin ^2}\alpha + \frac{{\sqrt 3 }}{2}\cos \alpha \sin \alpha \\
+ \frac{1}{4}{\cos ^2}\alpha + \frac{3}{4}{\sin ^2}\alpha - \frac{{\sqrt 3 }}{2}\cos \alpha \sin \alpha \\
= \frac{3}{2}{\cos ^2}\alpha + \frac{3}{2}{\sin ^2}\alpha \\
= \frac{3}{2}\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)\\
= \frac{3}{2}
\end{array}\]

LG c

\(\tan ({\pi  \over 3} - \alpha )\tan \alpha \tan ({\pi  \over 3} + \alpha ) \) \(= \tan 3\alpha \) (khi các biểu thức có ý nghĩa)

Ứng dụng: Tính tan10⁰ tan50⁰ tan110⁰

Lời giải chi tiết:

 Ta có:

\(\eqalign{
& \tan ({\pi \over 3} - \alpha )\tan \alpha \tan ({\pi \over 3} + \alpha ) \cr 
&  = \frac{{\tan \frac{\pi }{3} - \tan \alpha }}{{1 + \tan \frac{\pi }{3}\tan \alpha }}.\tan \alpha .\frac{{\tan \frac{\pi }{3} + \tan \alpha }}{{1 - \tan \frac{\pi }{3}\tan \alpha }}\cr &= {{\sqrt 3 - \tan \alpha } \over {1 + \sqrt 3 \tan \alpha }}\tan \alpha {{\sqrt 3 + \tan \alpha } \over {1 - \sqrt 3 \tan \alpha }} \cr 
& = {{3 - {{\tan }^2}\alpha } \over {1 - 3{{\tan }^2}\alpha }}\tan \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \cr 
& \tan 3\alpha = {{\tan 2\alpha + \tan \alpha } \over {1 - \tan 2\alpha .\tan \alpha }} \cr &= {{{{2\tan \alpha } \over {1 - {{\tan }^2}\alpha }} + \tan \alpha } \over {1 - {{2\tan \alpha } \over {1 - {{\tan }^2}\alpha }}\tan \alpha }}\,\, \cr 
& = \frac{{\frac{{2\tan \alpha  + \tan \alpha  - {{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 - {{\tan }^2}\alpha  - 2{{\tan }^2}\alpha }}{{1 - {{\tan }^2}\alpha }}}} \cr &= \frac{{3\tan \alpha  - {{\tan }^2}\alpha }}{{1 - 3{{\tan }^2}\alpha }}\cr &= {{3 - {{\tan }^2}\alpha } \over {1 - 3{{\tan }^2}\alpha }}.\tan\alpha \,\,\,\,\,\,(2) \cr} \)

Từ (1) và (2) suy ra điều phải chứng minh

Áp dụng:

\(\eqalign{
& tan{10^0}tan{50^0}tan{110^0} \cr&= \tan ({60^0} - {50^0})\tan {50^0}\tan ({60^0} + {50^0}) \cr 
& = \tan {150^0} = - \tan {30^0} = - {1 \over {\sqrt 3 }} \cr} \)

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