cos(-17兀/4)-sin(-17兀/4)=

cos(-17兀/4)-sin(-17兀/4)= 求证1/sin^2a+3/cos^2a>=4+2根号下31/sin²a +3/cos²a=(sin²a+cos²a)/sin²a +3(sin²a+cos²a)/cos²a=1+cos²a/sin²a+3+3sin²a/cos²a=4+cos²a/sin²a+3sin²a/cos²a≥4+2√[(cos sin（-26兀/3）.cos（-17兀/4）.tan（17兀/6）.利用公式求函数值 tan（-6分之5兀）+cos（-4分之23兀）+sin（-3分之17兀）计算 若x=兀/12 sin^4-cos^4= sin[兀/3 + arc cos(-1/4)] 的值 cos(-17π/4)-sin(-17π/4)=? cos(兀-2a)/sin(a-兀/4）=-跟号2/2求cos(a)+sin(a) 求值:sin(26π/3)+cos(-17π/4)= 已知cosα-sinα=3√2/5,若17兀/12 计算sin兀/8cos兀/8 sin兀/12×cos兀/12=? 为啥sinω+cosω =√2sin(ω+兀/4) 三角化简sin(x-5兀)cos[(-兀/2)-x]cos(8兀-x)/sin[x-(3兀/2)]sin(-x-4兀) 已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α）=根号2sin(θ-4π） 化简 cos(@-兀/2)/sin(5兀/2+@)*sin(@-2兀)*cos(2兀-@) 2cos+7tan兀-4sin二分之3兀-8cos二分之兀等于 2Sin(兀/4一阝)Cos(兀/4一阝)怎么=Cos^2阝一Sin^2阝的?求数学高手帮忙.