Question

化简根号1-sin1+根号（1-cos1）/2

Answer 1

∵0＜1/2＜π/4

∴0＜sin(1/2)＜cos(1/2)

∴√( 1-sin1 ) + √{ (1-cos1)/2 }

= √{ sin^2(1/2)+cos^2(1/2)-2sin(1/2)cos(1/2) } + √{ (1-[1-2sin^2(1/2)] }

= √{ sin(1/2)-cos(1/2) }^2 + √ sin^2(1/2)

= cos(1/2) - sin(1/2) + sin(1/2)

= cos(1/2)