Bài 113036

Question

Tính các giá trị của biểu thức
$1. P=\sin9^0.\sin 81^0$ $2. Q=\cos 10^0\cos50^0\cos70^0$
$3. R=\cos2^0\cos54^0\cos58^0\cos62^0\cos66^0$ $4.L=\tan 5^0\tan 55^0\tan 65^0$

score 0 · Answer 1

Ta có:
$1. P=\sin9^0\sin81^0=\sin9^0\cos9^0=\frac{\sin18^0}{2}$
Mà ta có: $\sin 54^0=\cos 36^0 \Leftrightarrow \sin3.18^0=\cos 2.18^0$
$\Leftrightarrow 3.\sin 18^0-4\sin^3 18^0=2\sin^2 18^0-1$
$4\sin^3 18^0-2\sin^2 18^0-3\sin 18^0+1=0              $
$\Leftrightarrow (\sin 18^0-1)(4\sin^2 18^0+2\sin 18^0-1)=0      $
Do $0<\sin 18^0<1$ nên $\sin 18^0=\frac{\sqrt{5}-1 }{4}$
Suy ra $P=\frac{\sqrt{5}-1 }{8}$
$2. 4Q=4\cos10^0\cos50^0\cos70^0$
     $=4\cos 10^0\cos(60^0-10^0)\cos(60^0+10^0)=\cos(3.10^0)=\cos 30^0$
Vậy $Q=\frac{1}{4}\cos 30^0=\frac{\sqrt{3} }{8}$
$3.R=\cos2^0.\cos54^0\cos58^0\cos62^0\cos66^0=\cos2^0.\cos58^0\cos62^0\cos54^0\cos66^0$
   $=[\cos2^0.\cos(60^0-2^0)\cos(60^0+2^0)]\cos54^0.\cos66^0$
   $=\frac{1}{4}\cos6^0.\cos54^0.\cos66^0=\frac{1}{4}.\cos6^0.\cos(60^0-6^0)\cos(60^0+6^0)\\=\frac{1}{4}.\frac{1}{4}\cos(3.6^0)$
   $\Leftrightarrow R=\frac{1}{16}\cos18^0$
Theo kết quả câu $1$ ta có $\cos 18^0=\sqrt{1-\sin^2 18^0}=\frac{\sqrt{10+2\sqrt{5} } }{4} \Rightarrow R=\frac{\sqrt{10}+2 \sqrt{5} }{64} $
$4. L=\tan5^0.\tan55^0.\tan65^0=\tan5^0\tan(60^0-5^0).\tan(60^0+5^0)\\=\tan (3.5^0)=\tan (45^0-30^0)$
    $=\frac{\tan45^0-\tan 30^0}{1+\tan 45^0.\tan 30^0}=\frac{1-\frac{1}{\sqrt{3} } }{1+\frac{1}{\sqrt{3} } }=\frac{\sqrt{3}-1 }{\sqrt{3}+1 }=\frac{(\sqrt{3}-1)^2 }{2}=2-\sqrt{3}$

Bài 113036

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Bài 113036

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