2sin^245°-tan330°×tan405°

2sin平方225度-cot330度*tan405度 怎么求! 高一数学sin(-1770)*cos1500+cos(-690)*sin780+tan405 化简下列各式：（1）a^2 sin(-1350度)+b^2 tan405度-（a-b）^2 cot765度-2ab co sin(-1740度)*cos(1470度)+cos(-660度)*sin(750度)*tan405度 解出过程 计算：sin²1°+sin²2°+sin²3°...+sin²45°+sin&# 数列求和 sin²1°+sin²2°+sin²3°+.+sin²88°+sin& sin²1°+sin²2°+sin²89°+sin²88°= sin²1°+sin²2°+……sin²88°+sin²89° sin^2 (0°)+sin^2( 1°)+.+sin^2(90du) 求和 求sin^2(44°)+sin^2(46°)+tan53°+tan37°的值 求证sin3°=（sin^2°-sin^1°）/sin1° 计算[3/(sin^2)20°-[1/(cos^2)20°]+64(sin^2)20°