三角函数的周期性与整除问题

青青子衿

\[\cos \left[ {{{\left( {{{10}^{2014}}} \right)}^ \circ }} \right]\sin {10^ \circ } - \sin \left[ {{{\left( {{{10}^{2014}}} \right)}^ \circ }} \right]\cos {10^ \circ } = ?\]

青青子衿

回复 1# 青青子衿
\[\begin{gathered}
  \sin \left[ {{{\left( {{2^n}} \right)}^ \circ }} \right] = \sin \left[ {{{\left( {{2^{n + 12}}} \right)}^ \circ }} \right]\left( {n \geqslant 2,n \in {N_ + }} \right) \\
  \sin \left[ {{{\left( {{3^n}} \right)}^ \circ }} \right] = \sin \left[ {{{\left( {{3^{n + 4}}} \right)}^ \circ }} \right]\left( {n \geqslant 2,n \in {N_ + }} \right) \\
  \sin \left[ {{{\left( {{5^n}} \right)}^ \circ }} \right] = \sin \left[ {{{\left( {{5^{n + 6}}} \right)}^ \circ }} \right]\left( {n \geqslant 1,n \in {N_ + }} \right)  \\
  \sin \left[ {{{\left( {{6^n}} \right)}^ \circ }} \right] =  - \sin \left[ {{{\left( {{6^{n + 1}}} \right)}^ \circ }} \right]\left( {n \geqslant 2,n \in {N_ + }} \right)  \\
  \sin \left[ {{{\left( {{7^n}} \right)}^ \circ }} \right] = \sin \left[ {{{\left( {{7^{n + 12}}} \right)}^ \circ }} \right]\left( {n \geqslant 1,n \in {N_ + }} \right)  \\
\end{gathered}\]