一道化简题

化简：{1-cos[(7π/2)-2α]+cos(2α-7π)}/{1+sin[2α+6π]+sin[(3π/2)+2α]}

热心网友

{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)=[2(sinα)^2-2sinαcosα]/[2(sinα)^2+2sinαcosα]=(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)。。

热心网友

{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)。。

热心网友

{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)。。