cos1°×cos1°+cos2°×cos2°+……+cosn
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cos1°×cos1°+cos2°×cos2°+……+cosncos1°×cos1°+cos2°×cos2°+……+cosncos1°×cos1°+cos2°×cos2°+……+cosn因为sin^2
cos1°×cos1°+cos2°×cos2°+……+cosn
cos1°×cos1°+cos2°×cos2°+……+cosn
cos1°×cos1°+cos2°×cos2°+……+cosn
因为sin^2+cos^2=1
cosA=sin(90-A)
原式=(n+1)/2+根号2/2
原式=cos?1+cos?2+…+cos?n = cos?1+cos?2+…cos?44+cos?45+sin?44+sin?43+…sin?2+sin?1 = (n-1)/2+cos45
cos1°×cos1°+cos2°×cos2°+……+cosn
cos1°+cos2°+cos3°.+co180°=?
cos1°+cos2°+cos3°+.+cos360°
cos1°+cos2°+cos3°+...+cos179°化简
cos1°+cos2°+cos3°+.+cos359°等于多少.
求证 1/(COS0°COS1°)+1/(COS1°COS2°)+...+1/(COS88°COS89°)=COS1°/(sin1°)^2裂项法
cos1°²+cos2°²+cos3°²+cos4°²...+cos89°²=?
cos1°+cos2°+cos3°+……+cos179°+cos180°的值是(
cos1°+cos2°+cos3°+...+cos²89°过程和答案!速度啊 多谢!
cos1°+cos2°+cos3°+...+cos180°的值为?到底是0还是-1?
cos1°+cos2°+cos3°+...+cos180°的值为?
Cos1°+cos2°+cos3°+……cos89°=
求cos1°+cos2°+cos3°+...+cos178°+cos179°的值（这里的数字都是度数比如1°）求cos1°+cos2°+cos3°+...+cos178°+cos179°的值
cos1°×cos1°+cos2°×cos2°+……+cos （nπ）×cos（nπ）=?（n∈正整数）o(∩_∩)o...Ps：不会打平方符号啊 有谁会 顺便教一下了
cos1°+cos2°+cos3°+……cos180°=_______谢谢大家,给出过程吧~~~~
cos2=a,cos1=?
cos0°+cos1°+.cos89°+cos90°等于多少
cos1°/sin44°-cos89°/sin136°=