Tính giá trị của biểu thức:
a) $A=\frac{\tan^2 \frac{\pi}{12} +\tan^2 \frac{5\pi}{12} }{\cot \frac{\pi}{24} } $             b) $B=\frac{\cos \frac{\pi}{7}.\cos \frac{4\pi}{7}. \cos \frac{5\pi}{7}}{\cos \frac{2\pi}{31}.\cos \frac{4\pi}{31}.\cos\frac{8\pi}{31}.\cos \frac{16 \pi}{31}.\cos \frac{32\pi}{31}} $


$A=\frac{M}{N} $,  ở đây  $M=\tan^2 \frac{\pi}{12}+\tan^2 \frac{5\pi}{12};  N=\cot \frac{\pi}{24}   $
* Ta tính $M$:
$M=(\frac{1}{\cos^2 \frac{\pi}{12} }-1 )+(\frac{1}{\cos^2 \frac{5\pi}{12} }-1 )= \frac{2}{1+\cos \frac{\pi}{6} }+\frac{2}{1+\cos \frac{5\pi}{6} }-2  $
     $=\frac{2}{1+\frac{\sqrt[]{3} }{2} }+\frac{2}{1-\frac{\sqrt[]{3} }{2} }-2=\frac{4(2-\sqrt[]{3} )}{4-3}+\frac{4(2+\sqrt[]{3} )}{4-3}-2=16-2=14    $
* Ta tính $N$:   $\cot \frac{\pi}{12}=\cot \frac{\pi}{6} +\frac{1}{\sin \frac{\pi}{6} }=\sqrt[]{3}+2   $
$N=\cot \frac{\pi}{24}=\cot \frac{\pi}{12}+\frac{1}{\sin \frac{\pi}{12} }=\cot \frac{\pi}{12}+\sqrt[]{1+\cot^2 \frac{\pi}{12} }     $
     $=\sqrt[]{3}+2+\sqrt[]{1+(2+\sqrt[]{3} )^2}=\sqrt[]{3}+2+\sqrt[]{8+4\sqrt[]{3} }=\sqrt[]{3}+2+\sqrt[]{(\sqrt[]{6}+\sqrt[]{2})^2}      $
Vậy $N=\sqrt[]{2}+\sqrt[]{3}+\sqrt[]{4}+\sqrt[]{6} $.   Suy ra   $A=\frac{14}{\sqrt[]{2}+\sqrt[]{3}+\sqrt[]{4}+\sqrt[]{6}    } $
b) $B=\frac{E}{F} $, ở đây   $E=\cos \frac{\pi}{7}.\cos \frac{4\pi}{7}. \cos \frac{5\pi}{7}   $,
                                 $F= \cos \frac{2\pi}{31}.\cos \frac{4\pi}{31}. \cos \frac{8\pi}{31}. \cos \frac{16\pi}{31}.\cos \frac{32\pi}{31}     $
* Tính $E$:   Vì   $\cos \frac{5\pi}{7}=\cos (\pi-\frac{2\pi}{7} )=-\cos \frac{2\pi}{7}   $
   nên   $\sin \frac{\pi}{7}E=-\sin \frac{\pi}{7}. \cos \frac{\pi}{7} . \cos \frac{2\pi}{7}. \cos \frac{4\pi}{7} = -\frac{1}{8}\sin \frac{8\pi}{7}=\frac{1}{8}\sin \frac{\pi}{7}          $
   Do đó   $E=\frac{1}{8} $.
* Tính $F$: Vì $\cos \frac{32\pi}{31}=\cos(\pi-\frac{\pi}{31} )=-\cos \frac{\pi}{31}  $  nên
                 $F=-\cos \frac{\pi}{31} .\cos \frac{2\pi}{31}.\cos \frac{4\pi}{31}. \cos \frac{8\pi}{31}. \cos \frac{16 \pi}{31}$.
Tương tự cách tính $E$, ta có :
   $\sin \frac{\pi}{31}.F=- \frac{1}{32} \sin \frac{32\pi}{31}=\frac{1}{32}\sin \frac{\pi}{31}$( vì  $\sin \frac{32\pi}{31}=\sin (\pi+\frac{\pi}{31} )=-\sin \frac{\pi}{31}  $)
 Do đó  $F=\frac{1}{32} $.  Vậy $B=\frac{E}{F} =\frac{\frac{1}{8} }{\frac{1}{32} }=4 $


viết xog mới thấy dài lê thê khà khà... –  rockoanh88 10-07-12 04:07 PM

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