Question

cos10°*cos20°*cos30°*cos40°*cos50°*cos60°*cos70°*cos80°​

Answer 1

Решение.

Упростить выражение . Применяем группировку множителей и формулы произведения косинусов

$\boldsymbol{cos\alpha \cdot cos\beta =\dfrac{1}{2}\, \Big(cos(\alpha +\beta )+cos(\alpha -\beta )\Big)}$      и  

$\boldsymbol{cos(180^\circ -\alpha )=-cos\alpha \ \ \Rightarrow \ \ cos\alpha =-cos(180^\circ -\alpha )}$  .    

$\bf cos10^\circ\cdot cos20^\circ \cdot cos30^\circ \cdot cos40^\circ \cdot cos50^\circ cos60^\circ \cdot cos70^\circ \cdot cos80^\circ =\\\\\\=(cos10^\circ \cdot cos80^\circ )\cdot (cos20^\circ \cdot cos70^\circ )\cdot (cos30^\circ cos60^\circ )\cdot (cos40^\circ \cdot cos50^\circ )=\\\\\\=\dfrac{1}{2^4}\Big(\underbrace{\bf cos90^\circ }_{0}+cos70^\circ \Big)\Big(cos90^\circ +cos50^\circ \Big)\Big(cos90^\circ +cos30^\circ \Big)\cdot \\\\\cdot \Big(cos90^\circ +cos10^\circ \Big)=$

$\bf =\dfrac{1}{2^4}\cdot cos70^\circ \cdot cos50^\circ \cdot cos30^\circ \cdot cos10^\circ =\\\\\\=\dfrac{1}{16}\cdot (cos70^\circ \cdot cos10^\circ )\cdot (cos50^\circ \cdot cos30^\circ )=\\\\\\=\dfrac{1}{16}\cdot \dfrac{1}{2^2}\cdot \Big(cos80^\circ +cos60^\circ \Big)\cdot \Big(cos80^\circ +cos20^\circ \Big)=$  

$\bf =\dfrac{1}{64}\cdot \Big(cos80^\circ \cdot cos80^\circ +cos80^\circ \cdot cos20^\circ +cos60^\circ \cdot cos80^\circ +cos60^\circ \cdot cos20^\circ\Big)=\\\\\\=\dfrac{1}{64}\cdot \dfrac{1}{2}\cdot \Big(cos160^\circ +cso0^\circ +cos100^\circ +cos60^\circ +cos140^\circ +cos20^\circ +\\\\+cos80^\circ +cos40^\circ \Big)=\\\\\\=\dfrac{1}{128}\cdot \Big(-cos20^\circ +1-cos80^\circ +cos60^\circ -cos40^\circ +cos20^\circ +cos80^\circ +cos40^\circ \Big)=$

$\displaystyle \bf =\dfrac{1}{128}\cdot \Big( 1+cos60^\circ \Big)=\frac{1}{128}\cdot \Big(1+\frac{1}{2}\Big)=\frac{1}{128}\cdot \frac{3}{2}=\frac{3}{256}$