sin^2(42°+α)+cot(25°+β)cot(β-65°)+sin^2(48°-α）=？？？

显示全部楼层 · 2011-9-12 10:22:07

我来试试吧...解：sin^2(42°+α)+sin^2(48°-α)=sin^2(42°+α)+cos^2[90-(48°-α)]
=sin^2(42°+α)+cos^2(42°+α)=1
cot(25°+β)cot(β-65°)=cot(25°+β)tan[90-(β-65°)]=cot(25°+β)tan(25°+β)=1
所以原式=1-1=0...

千问 · 2011-9-12 10:22:07

sin^2(42°+α)+cot(25°+β)cot(β-65°)+sin^2(48°-α）=sin^2(42°+α)+sin^2(48°-α）+cot(25°+β)cot(β-65°)=sin^2(42°+α)+cos2（42°+a）+cot（25°+β）[-tan（25°+β）]=1-1=0【cot(β-65°)=-tan（β-...