填充第五題
\( \displaystyle cos 36^{\circ}=\frac{\sqrt{5}+1}{4} \)
\( cos 2 \theta=2cos^2 \theta-1 \)　，　\( cos 72^{\circ}=2cos^2 36^{\circ}-1=\frac{\sqrt{5}-1}{4} \)

\( \displaystyle tan^2 18^{\circ} tan^2 54^{\circ}=\frac{sin^2 18^{\circ}}{cos^2 18^{\circ}} \times cot^2 36^{\circ}=\frac{sin^2 18^{\circ}}{cos^2 18^{\circ}} \times \frac{cos^2 36^{\circ}}{sin^2 36^{\circ}}=\frac{\frac{1-cos 36^{\circ}}{2}}{\frac{1+cos 36^{\circ}}{2}}\times \frac{\frac{1+cos 72^{\circ}}{2}}{\frac{1-cos 72^{\circ}}{2}}=\frac{1-cos 36^{\circ}}{1+cos 36^{\circ}} \times \frac{1+cos 72^{\circ}}{1-cos 72^{\circ}}=\frac{1}{5} \)

[ 本帖最後由 shingjay176 於 2014-6-18 01:09 PM 編輯 ]

請教各位老師~填充第10題可否用根與係數?

填充題第十題
\[\begin{array}{l}
2{x^4} - 7{x^3} + 6{x^2} + 4x + 1 = 2\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \gamma } \right)\left( {x - \delta } \right)\\
\\
\;\;\;\left( {2 - {\alpha ^2}} \right)\left( {2 - {\beta ^2}} \right)\left( {2 - {\gamma ^2}} \right)\left( {2 - {\delta ^2}} \right)\\
= \left( {\sqrt 2  - \alpha } \right)\left( {\sqrt 2  - \beta } \right)\left( {\sqrt 2  - \gamma } \right)\left( {\sqrt 2  - \delta } \right)\left( {\sqrt 2  + \alpha } \right)\left( {\sqrt 2  + \beta } \right)\left( {\sqrt 2  + \gamma } \right)\left( {\sqrt 2  + \delta } \right)\\
= \frac{1}{2}\left\{ {2{{\left( {\sqrt 2 } \right)}^4} - 7{{\left( {\sqrt 2 } \right)}^3} + 6{{\left( {\sqrt 2 } \right)}^2} + 4\sqrt 2  + 1} \right\} \times \frac{1}{2}\left\{ {2{{\left( { - \sqrt 2 } \right)}^4} - 7{{\left( { - \sqrt 2 } \right)}^3} + 6{{\left( { - \sqrt 2 } \right)}^2} + 4\left( { - \sqrt 2 } \right) + 1} \right\}\\
= \frac{{241}}{4}
\end{array}\]

[ 本帖最後由 shingjay176 於 2014-6-18 08:58 PM 編輯 ]

用根與係數做，你會轉非常大圈，才走到終點算出答案。