大师作文网πD2 V/4
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/18 14:03:23
cos(π/12)^4+sin(π/12)^4=(cos²(12\π)+sin²(12\π))²-2sin²(12\π)cos²(12\π)=1-(√
sin(π/4-x)=sin(π/4)cosx-cos(π/4)sinx=（√2/2）cosx-（√2/2）sinx;cos(x+π/4)=cosxcos(π/4)-sinxsin(π/4)=（√2/
原式=sin(a-π/4)+sin[π/2-(a+π/4)]=sin(a-π/4)+sin(π/4-a)=sin(a-π/4)-sin(a-π/4)=0.
y=tan2xtanx^3=2tanx/(1-tan²x)•tanx^3=2tanx^4/(1-tan²x)=2tanx^4/(1-tan²x)=[2(tan
因为sin2θ=a=2sinθcosθ=acosθsinθ=a/2sin（π/4-θ）=√2/2cosθ-√2/2sinθ所以sin^2（π/4-θ）=1/2(cosθ-sinθ)^2=1/2(1-2
弄清楚kπ+π4≤x再问：那怎么知道π/4和-1谁比较大啊再答：π/4当然大于-1，是说π/4和1吧，π=3.14....,π/4
cosπ/4cosπ/3-sinπ/4sinπ/3=cos45°cos60°-sin45°sin60°=(√2/2)*(1/2)-(√2/2)*(√3/2)=(√2-√6)/4
cos2(π/4-α)-sin2(π/4-α)=cos(π/2-2α）-sin(π/2-2α)=sin2α-cos2α=√2(√2/2sin2α-cos2α)=√2sin(2α-π/4)根据二倍角公式
sin(π/4-α)+cos(π/4+α)=sin(π/4-α)+cos[π/2-(π/4-α)]=sin(π/4-α)+sin(π/4-α)=2sin(π/4-α)
1.12.±33.两面都平方x²=3∴x＝±√34.∵x²＝﹙－3.7﹚²＝3.7²∴x＝±3.75.由题可知﹙分配律﹚﹙√3-1﹚·√2＝0.732×1.41
π/4
1)π/4
sin(π/4-a)=cos(π/2-(π/4-a))=cos(π/4+a)cos(a-π/4)=cos(π/4-a)∴原式=cos(π/4+a)cos(π/4-a)由积化和差得原式=1/2(cos(
sin(π+π/6)-cos(π+π/4)cos(-π/2)+1=-sin(π/6)+cos(π/4)cos(+π/2)+1=-1/2+cos(π/4)*0+1=1/2
/π-5/+/4-π/+/-π+3.1/=5-π+4-π+π-3.1=5.9-π如还有新的问题,请不要追问的形式发送,另外发问题并向我求助或在追问处发送问题链接地址,
cosπ/3+tanπ/4+3tan²π/6+sinπ/2+cosπ+sin3π/2=1/2+1+3*(根号3/3)²+1-1-1=1/2+3*(1/3)=1/2+1=3/2
原式=sin(4π+π/6)+cos(6π+π/4)-tan(6π+π/4)-sin(8π+2π/3)=sinπ/6+cosπ/4-tanπ/4-sin2π/3=1/2+√2/2-1-√3/2=（√2
∵tan（a+π4）=tana+11−tana=13∴tana=-12因此，(sina−cosa)2cos2a=sin2a−2sinacosa+cos2acos2a−sin2a分子分母都除以cos2a
3π/4
由于π4-a是第一象限角，∴sin（π4-a）=513，∴sin(π2−2a)sin(π4+a)=sin2(π4−a)cos(π4−a)=2sin（π4-a）=1013．