(π-1)^0+(1/2)^-1+|5-√27|-2√3=?

两道计算题1、5^-2-(-4)^-2+(π-1)^0 2、(-1/3)^0+(-1/3)^-1+(-1/3)^-2 （1)已知5π/2 (π-1)^0+(-1/2)^-1+|5-根号27|-2根号3 计算 (π-1)^0+(1/2)^-1+5-根号27的绝对值-2根号3 计算题 (π-1)^0+(1/2)^-1+|5-√27|-2√3=? sin0=0 sin(1/6π)=1/2sin(1/3π)=√3/2sin(1/2π)=1sin(2/3π)=√3/2sin(5/6π)=1/2sin(π)=0sin(-π)=0sin(-5/6π)=-1/2sin(-2/3π)=-√3/2sin(-1/2π)=-1sin(-1/3π)=-√3/2sin(-1/6π)=-1/2sin(7/6π)=-1/2sin(4/3π)=-√3/2sin(5/3π)=-√3/2sin(11/6π)=-1/ cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0 1-11^2+(2/1)^-1-(200-π）^0 α∈(0,π),已知sin(3π-α)-cos(α-9π)=1/5,则1/tan(2π-α)=讲讲a∈（0, 计算:根号4+(π-2)^0-|-5|+(-1)^2012+(1/3)^-2 |-2|+(π-3)^0-(1/3)^0+(-1)^2010 |-2|+(-1)-(π-√3)0次方 10³+（1/30）^-2×（π-5）^0-（-3）³×0.3^-1+｜-23｜； 求证cosπ/5 + cos3π/5=1/2 若x∈(0,π) ,且sin(π/2+x)+sin(π-x)=1/5,则tanx的值 若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=?若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα= 已知0小于A小于2/π,若COSπ-SINπ=-根号5/5,试求 2SINACOSA-COSA+1/1-TANA的值 (π-)^0-(1/2)^-1+(2/3)^2008*(-1.5)^2009