搜搜考试网sin(四分之π x)sin(四分之π-x)=六分之一
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sinα=-√2/4α=-20.7°{α|α=k×360°-20.7°,k∈Z}
∵sin(π/4-x)=12/13∴cos(π/4+x)=cos(π/2-(π/4-x))=sin(π/4-x)=12/13∵sin(π/4-x)=(cosx-sinx)/√2∴cosx-sinx=√
有sin(x-45°)=√2／4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3／8.tanx+1／tanx
∵x∈(-π/4,0)∴x+π/4∈(0,π/4)∵cos(x+π/4)=4/5∴sin(x+π/4)=3/5sinx=sin(x+π/4-π/4)=sin(x+π/4)cosπ/4-cos(x+π/
f(x)=sin^2x+√3sinxcosx=1-cos²x＋√3／2sin2x=1-（1＋cos2x）／2＋√3／2sin2x＝1／2＋√3／2sin2x－1／2cos2x=1／2＋sin
f(x)=sin2x+2sin（π/4-x）cos（π/4-x）=sin2x+cos2x=√2sin（2x+π/4）所以T=2π/2=πx属于[-π/12,π/2]得到2x+π/4属于[π/12,5π
∵0＜x＜π/4∴0再问：求的是cos2x除以cos（四分之派加x)的值，你算的是cos2x除以cos（四分之派减x)，算错了，正确答案是13分之24，帮忙再算一下，谢谢！再答：啊我算的是cos2x/
∫(-π/4,π/4)(x^3+cosx)/[1+sin^(2)x]dx=∫(-π/4,π/4)x^3/[1+sin^(2)x]dx+∫(-π/4,π/4)cosx/[1+sin^(2)x]dx（前者
sin(2x-π/4)>0且求sin(2x-π/4)的增区间即可2kπ
y=sin(π/4—2X)=-sin(2x-π/4)的单调增区间即为sin(2x-π/4)的单调减区间,由π/2+2kπ≤2x-π/4≤3π/2+2kπ3π/4+2kπ≤2x≤7π/4+2kπ3π/8
sin四分之派cos四分之派=1/2sinπ/2=1/2
你好!数学之美团为你解答α,β∈(3π/4,π)α+β∈(3π/2,2π),β-π/4∈(π/2,3π/4)sin(α+β)=-3/5,sin(β-π/4)=12/13∴cos(α+β)=4/5,co
cos45°+sin30°=√2/2+1/2=（√2+1）/2
由两角和的正、余弦公式得y=√2cos2x,故单调减区间为2kπ≤2x≤(2k+1)π,单调增区间(2k-1)π≤2x≤2kπ,也即减区间[kπ,kπ+π/2],增区间[kπ-π/2,kπ].对称轴为
∵0＜a＜四分之π＜β＜四分之三π∴3π/4
28°25′
17π/12＜x＜7π/4,得5π/3＜x+π/4＜2πcos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)
原式=cos[2(π/4-a)]=cos(π/2-2a)=sin2a