[(1+sinx)/cosx]*[sin2x/2cos²(π/4-x/2)]
问题描述:
[(1+sinx)/cosx]*[sin2x/2cos²(π/4-x/2)]
答
=[(1+sinx)/cosx]*{sin2x/[cos(π/2-x)+1]}=[(1+sinx)/cosx]*[2sinxcosx/(sinx+1)]=2sinx
[(1+sinx)/cosx]*[sin2x/2cos²(π/4-x/2)]
=[(1+sinx)/cosx]*{sin2x/[cos(π/2-x)+1]}=[(1+sinx)/cosx]*[2sinxcosx/(sinx+1)]=2sinx