由Sin(A+B)=SinACosB+CosASinB
Sin(A-B)=SinACosB-CosASinB
则Sin(A+B)+Sin(A-B)=2SinACosB
Cos1°+cos2°+cos3°+……cos89°
=2Sin1°(Cos1°+cos2°+cos3°+……cos89°)/2Sin1°
=(2 Sin1°Cos1°+2 Sin1°Cos2°+2 Sin1°Cos3°+•••+2 Sin1°Cos88°+2 Sin1°Cos89°)/ 2Sin1°
=(Sin2°+ Sin0°+ Sin3°+ Sin(-1)°+ Sin4°+ Sin(-2)°+•••+ Sin89°+ Sin(-87)°+ Sin90°+ Sin(-88)°)/ 2Sin1°
=(Sin2°+ Sin0°+ Sin3°- Sin1°+ Sin4°- Sin2°+•••+ Sin89°- Sin87°+ Sin90°- Sin88°)/ 2Sin1°
=( Sin90°- Sin1°)/ 2Sin1°
=(1- Sin1°)/ 2Sin1°
Cos1°+cos2°+cos3°+……cos89°=
Cos1°+cos2°+cos3°+……cos89°=
数学人气:165 ℃时间:2020-02-01 09:34:48
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