一道化简题化简：{1-cos[(
{1-cos[(7π/2)-2α]+cos(2α-7π)}/{1+sin[2α+6π]+sin[(3π/2)+2α]}
={1-cos[3π+(π/2)-2α]+cos(7π-2α)}/(1+sin2α-cos2α)
={1-cos[(π/2)-2α]+cos(6π+π-2α)}/(1+sin2α-cos2α)
={1-sin2α+cos(π-2α)}/(1+sin2α-cos2α)
={1-sin2α-cos2α}/(1+sin2α-cos2α)
=[(sinα)^2+(cosα)^2-2sinαcosα-(cosα)^2+(sinα)^2]/ [(sinα)^2+(cosα)
^2+...全部
{1-cos[(7π/2)-2α]+cos(2α-7π)}/{1+sin[2α+6π]+sin[(3π/2)+2α]}
={1-cos[3π+(π/2)-2α]+cos(7π-2α)}/(1+sin2α-cos2α)
={1-cos[(π/2)-2α]+cos(6π+π-2α)}/(1+sin2α-cos2α)
={1-sin2α+cos(π-2α)}/(1+sin2α-cos2α)
={1-sin2α-cos2α}/(1+sin2α-cos2α)
=[(sinα)^2+(cosα)^2-2sinαcosα-(cosα)^2+(sinα)^2]/ [(sinα)^2+(cosα)
^2+2sinαcosα-(cosα)^2+(sinα)^2]
=[2(sinα)^2-2sinαcosα]/[2(sinα)^2+2sinαcosα]
=[2sinα(sinα-cosα)]/ [2sinα(sinα+cosα)]
=(sinα-cosα)/(sinα+cosα)
=(sinα/cosα-cosα/cosα)/(sinα/cosα+cosα/cosα)
=(tanα-1)/(tanα+1)
=[tanα-tan(π/4)]/[1+tanαtan(π/4)]
=tan(α-π/4)。
。收起
