化简((1+sinx+cosx)(sinx/2-cosx/2))/根号(2+2cosx)(0
问题描述:
化简((1+sinx+cosx)(sinx/2-cosx/2))/根号(2+2cosx)(0
答
((1+sinx+cosx)(sinx/2-cosx/2))/根号(2+2cosx)
=2(sinx/2*cosx/2+cox^2x/2)(sinx/2-cosx/2))/(2cosx/2)
=2cosx/2*(sinx/2+coxx/2)(sinx/2-cosx/2))/(2cosx/2)
=sin^2(x/2)-cos^2(x/2)
=-cosx
答
cosx=2cos^2(x/2)-1
因为0