まず、0＜sin(\(\frac{1}{9}\))π＜1,0＜cos(\(\frac{1}{9}\))π＜1
①
cos(\(\frac{7}{18}\))π=cos{(\(\frac{1}{2}\))π-(\(\frac{1}{9}\))π}=sin(\(\frac{1}{9}\))π=a
②
sin(\(\frac{25}{18}\))π=sin{π+(\(\frac{7}{18}\))π}=-sin(\(\frac{7}{18}\))π
=-sin{(\(\frac{1}{2}\))π-(\(\frac{1}{9}\))}π=-cos(\(\frac{1}{9}\))π=-\(\sqrt{\quad}\)(1-\(a^{2}\))
③
tan(\(\frac{29}{18}\))π=tan{(\(\frac{3}{2}\))π+(\(\frac{1}{9}\))π}=-cot(\(\frac{1}{9}\))π=-{\(\sqrt{\quad}\)(1-\(a^{2}\))}/a
④
cos(-\(\frac{11}{18}\))π=cos(\(\frac{11}{18}\))π=cos{π-(\(\frac{7}{18}\))π}=-cos(\(\frac{7}{18}\))π=-a