1π2π3π4π5π······100π各是多少?（π为3.14）1π2π3π4π5π······40π即可

1π2π3π4π5π······100π各是多少?（π为3.14）1π2π3π4π5π······40π即可 sin3分之4π·cos6分之5π·tan(-3分之4π)= sin2π/3· cos3π/2· tanπ/4= 若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根：±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=（1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用 根号sin^α-sin^4α（π/2〈α〈π）化简根号sin平方α-sin4次方α（π/2〈α〈π）化简·········· 已知函数y=3cos^2x+2√3sinxcosx+sin^2x,求(1)当x属于R时,函数的最大值和最小值；（2）当x属于[-π/4,π/4]时,函数的最大值和最小值.求求大家帮帮我好嘛好嘛好嘛························· 化简：cos(A-π/2）·cot（-A-3π/2）·色彩（-A+5π/2）·tan（5π/2+A）, sin(-11π/6)+cos12π/5·tan3π+cos(-π/4) 化简(2×√3-π)÷√2(保留3个有效数字) 大哥大姐,帮帮忙啊··················化简，你那个我也会啊，真是············ 化简:sin(a+5π)cos(-π/2-a)·cos(8π-a)/sin(a-3π)·sin(-a-4π)化简:sin(a+5π)cos(-π/2-a)·cos(8π-a)/sin(a-3π/2)·sin(-a-4π) tan(π+α)=-1/2求sin(α-7π)·cos(α+5π) 一道小数学题说原因··········已知cos（π/2+φ）=√3/2,且/ φ/ 化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)RT、quickly~ 设f(x)=sin(πx/2+π/3)+cos(πx/2+π/6),x属于z,求f(1)+f（2）+······+f（2007）的值 π=3.14······后面的数?我只知道π=3.1415926535897932······, 三角函数···cos(π/5 +x) = -1/2 ,怎么化简? 求值sin(π/3+arctan1/2)最好要有过程···· 求不定积分∫(πx^3+1)^4·3πx^2dx