Question

(9 класс)
1. вычислить:
128sin²20˚ sin²40˚ sin²60˚ sin²80˚

Answer 1

$128\cdot \sin^220\textdegree \sin^240\textdegree \sin^260\textdegree \sin^280\textdegree =\\=128\cdot \sin^2 60\textdegree \cdot \Big(\sin 20\textdegree \sin 40\textdegree \sin 80\textdegree\Big)^2 =\\=128\cdot \bigg(\dfrac{\sqrt3}2\bigg)^2 \cdot \Big(\sin 20\textdegree \sin \big(60\textdegree -20\textdegree\big) \sin \big(60\textdegree +20\textdegree\big) \Big)^2 =$

$=128\cdot \dfrac34 \cdot \Big(\sin 20\textdegree\cdot \big( \sin 60\textdegree \cdot \cos 20\textdegree-\sin 20\textdegree \cdot \cos 60\textdegree\big) \times\\~~~~~~~~~~~~~~~~~~~~~~~~ \times \big( \sin 60\textdegree \cdot \cos 20\textdegree+\sin 20\textdegree \cdot \cos 60\textdegree\big)\Big)^2 =$

$=32\cdot 3 \cdot \bigg(\sin 20\textdegree\cdot \Big( \big(\sin 60\textdegree \cdot \cos 20\textdegree\big)^2-\big(\sin 20\textdegree \cdot \cos 60\textdegree\big)^2 \Big) \bigg)^2 =\\=32\cdot 3 \cdot \bigg(\sin 20\textdegree\cdot \Big( \bigg(\dfrac {\sqrt3}2\bigg)^2 \cdot \cos^2 20\textdegree-\bigg(\dfrac 12\bigg)^2 \cdot \sin^2 20\textdegree \Big) \bigg)^2 =$

$=32\cdot 3 \cdot \dfrac 1{16}\bigg(\sin 20\textdegree\cdot \Big(3\cos^2 20\textdegree-\sin^2 20\textdegree \Big) \bigg)^2 =\\=6\bigg(\sin 20\textdegree\cdot \Big(3\big(1-\sin^2 20\textdegree\big)-\sin^2 20\textdegree \Big) \bigg)^2 =\\=6\bigg(\sin 20\textdegree\cdot \Big(3-4\sin^2 20\textdegree\Big) \bigg)^2 =\\=6\bigg(3\sin 20\textdegree-4\sin^3 20\textdegree \bigg)^2 =6\bigg(\sin \big(3\cdot 20\textdegree\big)\bigg)^2=\\=6\Big(\sin 60\textdegree\Big)^2=6\cdot\bigg(\dfrac{\sqrt3}2\bigg)^2=\boldsymbol{4,5}$

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Использованы формулы

$\sin \big(\alpha \pm \beta \big)=\sin \alpha \cos \beta \pm \sin \beta \cos \alpha \\\sin^2\alpha +\cos^2\alpha =1\\(a-b)(a+b)=a^2-b^2\\3\sin \alpha -4\sin^3 \alpha =\sin\big(3\alpha \big)$