Cos1°+cos2°+cos3°+……cos89°=
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Cos1°+cos2°+cos3°+……cos89°=Cos1°+cos2°+cos3°+……cos89°=Cos1°+cos2°+cos3°+……cos89°=由Sin（A+B）=SinACosB+
Cos1°+cos2°+cos3°+……cos89°=
Cos1°+cos2°+cos3°+……cos89°=
Cos1°+cos2°+cos3°+……cos89°=
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°
乘以sin0.5
0.5*(sin1.5-sin0.5)+0.5*sin(sin2.5-sin1.5)+...+0.5*(sin89.5-sin88.5)=0.5*(sin89.5-sin0.5)
原式=0.5*cot0.5-0.5
原式中不是求平方和哈，那么第二种做法好像有问题哟！
cos1°+cos2°+cos3°+……+cos179°+cos180°的值是(
Cos1°+cos2°+cos3°+……cos89°=
cos1°+cos2°+cos3°.+co180°=?
cos1°+cos2°+cos3°+.+cos360°
cos1°+cos2°+cos3°+...+cos179°化简
cos1°+cos2°+cos3°+.+cos359°等于多少.
cos1°+cos2°+cos3°+……cos180°=_______谢谢大家,给出过程吧~~~~
cos1°×cos1°+cos2°×cos2°+……+cosn
求cos1+cos2+cos3+…+cos178+cos179的值
(sin1°-cos1°)(sin2°-cos2°)(sin3°-cos3°)……（sin89°-cos89°）=一定要告诉我原因啊 谢谢最后得0.
cos1°的二次方+cos2°的二次方+cos3°的二次方+……cos89°的二次方=
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cos1°+cos2°+cos3°+...+cos²89°过程和答案!速度啊 多谢!
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