化简并计算sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌)

化简并计算sina(丌/6+2k丌).cos(11/3丌)-tan(5/4丌+2k丌) 求值tan(kπ+π/6)cos(kπ-4π/3),k属于z 化简sin(-2/3pai+kpai)cos(pai/6+kpai)tan(pai/4+kpai),k属于z cosπ/3－tanπ/4+cosπ－sin3π/2 化简计算 sina+cosa=-3/根号5,且|Sina|>|cos a|,求tan(a/2) 化简:sin(3丌-a)/tan(a-5丌)x tan(-a+2丌)/cot(丌-a)x cos(4丌-a)/sin(- 已知sina(2a+b)=3sinb,a不等于kπ+π/2,a+b不等于kπ+π/2,k为正整数,求证：tan(a+b) [k(k+1)/12](3k^2+11k+10)+(k+1)(k+2) 怎么化简 计算2/（k+1）（k+3) +2/（k+3)（k+5)+…+2/(k+2003)(k+2005) 怎么证明tan(π/(2k+1))×tan（2π/(2k+1))×tan(3π/(2k+1))×.tan(k/(2k+1 有人懂不,已知tan(3.14-A)=-3求：1、4sinA-2cosA/5cosA+3sinA的值2、sinA*cos 已知tana=3,计算 1、4sina-2cosa/5cosa+3sina 2、sinacosa 3、(sina+cos