设a∈(3π/2,2π)化简根号下1/2+1/2根号下1/2+1/2cos2a
问题描述:
设a∈(3π/2,2π)化简根号下1/2+1/2根号下1/2+1/2cos2a
答
解析:
因为a∈(3π/2,2π),所以:cosa>0
则:1/2+(1/2)*√(1/2+1/2cos2a)
=1/2+(1/2)*√[(1/2)*(1+cos2a)]
=1/2+(1/2)*√[(1/2)*(2cos²a)]
=1/2+(1/2)*cosa
=1/2*(1+cosa)
=1/2*[2cos²(a/2)]
=cos²(a/2)