ข้อ4)
$\frac{tanA}{1-tan^{2}A}=sin^{2}20^{\circ}-sin160^{\circ }sin220^{\circ }+sin^{2}320^{\circ } $
$\frac{tan2A}{2}= sin^{2}20^{\circ}+sin20^{\circ }sin40^{\circ }+sin^{2}40^{\circ } $
$tan2A=2sin^{2}20^{\circ}+2sin20^{\circ }sin40^{\circ }+2sin^{2}40^{\circ } $
$tan2A=1-cos40^{\circ }+cos20^{\circ }-cos60^{\circ }+1-cos80^{\circ }$
$tan2A=\frac{3}{2} +cos20^{\circ }-cos40^{\circ }-cos80^{\circ }$
$tan2A=\frac{3}{2}+2sin30^{\circ }sin10^{\circ }-sin10^{\circ }$
$\therefore tan2A=\frac{3}{2}$
และจากสูตร $tan3\theta =\frac{3tan\theta -tan^{3}\theta }{1-3tan^{2}\theta } $
จะได้ว่า ..............
$tan6A=\frac{3tan2A -tan^{3}2A}{1-3tan^{2}2A } $
$tan6A=\frac{(3)(\frac{3}{2}) -(\frac{3}{2})^{3} }{1-(3)(\frac{3}{2})^{2} } $
$tan6A=\frac{\frac{9}{2} -\frac{27}{8} }{1-\frac{27}{4} } $
$tan6A=-\frac{9}{46} $
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