Bài 111730

Question

Tính giá trị của biểu thức sau:
a) $\frac{2\cos \frac{\pi}{5}.\cos \frac{2\pi}{5}}{\tan \frac{\pi}{24} +\tan \frac{7\pi}{24}} $
b) $\frac{2007}{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}} $
c) $1-\frac{2}{3} \sin^2 10^0. \sin^2 50^0. \sin^2 70^0 $

score 0 · Answer 1

a) * Tính $A=2\cos \frac{2\pi}{5} \cos \frac{\pi}{5} $
                $\sin \frac{\pi}{5}. A= \frac{1}{2} \sin \frac{4\pi}{5} \Rightarrow A=\frac{1}{2}$
   * Tính $B =\tan \frac{\pi}{24} +\tan \frac{7\pi}{24} = \frac{\sin \frac{\pi}{24} }{\cos \frac{\pi}{24} } + \frac{\sin \frac{7\pi}{24} }
{\cos \frac{7\pi}{24} }=\frac{\sin \frac{\pi}{3} }{\cos \frac{\pi}{24}. \cos \frac{7\pi}{24} }=\frac{2\sqrt[]{3} }{\sqrt[]{2}+1 } $
                   $=2(\sqrt

[]{6}-\sqrt[]{3} )$
Vậy $\frac{2\cos \frac{\pi}{5}.\cos \frac{2\pi}{5} }{\tan \frac{\pi}{24} +\tan \frac{7\pi}{24} }=\frac{1}{4(\sqrt[]{6}-\sqrt[]{3})}=\frac

{\sqrt[]{6}+\sqrt[]{3}}{12}   $
b)* Tính $A=\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}   $
               $2\sin \frac{\pi}{7}.A=2\cos \frac{2\pi}{7}. \sin \frac{\pi}{7}+2\sin \frac{\pi}{7} \cos \frac{4\pi}{7}=2\sin \frac{\pi}{7} \cos \frac{6
\pi}{7}=-\sin \frac{\pi}{7}        $
               $ \Rightarrow A=-\frac{1}{2} $
Do đó $\frac{2007}{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos\frac{6\pi}{7}}=\frac{2007}{\frac-{1}{2} } =-4014 $
c) * Tính $C=\sin 10^0. \sin 50^0. \sin 70^0=\frac{1}{2} \sin 10^0 (\cos 20^0-\cos 120^0)$
                $=\frac{1}{2}\sin 10^0 \cos 20^0 + \frac{1}{4}\sin 10^0 $
                $=\frac{1}{4}(\sin30^0-\sin10^0) + \frac{1}{4} \sin 10^0 =\frac{1}{4} \sin 30^0   $
                $\Rightarrow C=\frac{1}{8} $
Vậy $1-\frac{2}{3} C^2=1- \frac{2}{3}. \frac{1}{64}=1-\frac{1}{3.32}= 1-\frac{1}{96}=\frac{95}{96}      $

Bài 111730

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Bài 111730

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