∫(上5π 4下π 4)1 cos²xdx

√[1-2sin(π+4)cos(π+4)]√[sin²(π+4)-2sin(π+4)cos(π+4)+cos²(π+4)]=√{[sin(π+4)-cos(π+4)]²

根据:cos（a-π/6）+sina=（4*根号3）/5有:cosacosπ/6+sinasinπ/6+sina=（4*根号3）/5(根号3)/2cosa+3/2sina=（4*根号3）/5得到:co

根据这个算算你就知道了http://upload.wikimedia.org/math/c/a/c/cac95c43d3831cd1b61dedeb81dec77e.png

[sinα+2cos(5π/2+α)]/[cos(π-α)-sin(π/2-α)]=-1/4即(sinα-2sinα)/(-cosα-cosα)=-1/4,所以sinα/(2cosα)=-1/4,所以

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

sin(θ+kπ)=-2cos(θ+kπ),可得tanQ=-24sinθ-2cosθ/5cosθ+3sinθ（分子分母同时除以cosQ）=10⑵(1/4)sin平方θ+(2/5)cos平方θ（分子分母

∫(sin^3x)/(cos^4x+sin^2x)dx=-∫(sin^2x)/(cos^4x+sin^2x)dcosx=-∫(1-cos^2x)/(cos^4x+1-cos^2x)dcosx令cosx

√[1-cos(2π+θ)]/[1+cos(2π+θ)]+√[[1+cos(2π+θ)]/[1-cos(2π+θ)]=√(1-cosθ)/(1+cosθ)+√(1+cosθ)/(1-cosθ)=√(1

(cosa)^2=(cos2a+1)/2(cosa)^2=(cos2a+1)^2/4cos(π/8))^4=(cosπ/4+1)^2/4=[3+2√2]/8cos(3π/8))^4=(cos3π/4+

1-sin2a-cos^2(a-π/4)=sin^2(a-π/4)-sin2a=(根号2sina-根号2cosa)^2-sin2a=2(cosa^2-2sinacosa+cosa^2)-sin2a=2

=cosα+cos2π/3*cosα-sin2π/3*sinα+cos4π/3*cosα-sin4π/3*sinα=cosα-1/2cosα-（根号3）/2sinα-1/2cosα+（根号3）/2si

cos(2x-π/4)>=02kπ-π/2

cos(－19π/4)=ccos(19π/4)=cos[4π＋3π/4]=cos(3π/4)=cos[π－π/4]=－cos(π/4)=－√2/2

2cosα再问：怎么来的？再答：1=cos²α＋sin²α在2kπ-π/4≤α≤2kπ+π/4下，cosα＞sinα再问：为什么cosα＞sinα？再答：周期是2kΠ,在-π/4≤

√(1-cosx)(1+cosx)+√(1+cosx)(1-cosx)=2√(1-cosx)(1+cosx)=2√(1-cosx^2)=2√(Sinx^2)x∈（π/2,π）=2Sinx

原式=√(sin²4)=|sin4|=-sin4

[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2

把它展开就为cos^2x+x^3cos^2x的定积分,因为后一部分为奇函数直接消掉积分出来就是0,则只有cos^2x的积分,化成（cos2x+1）/2的积分,为偶函数,直接就是0到π上的积分的两倍,解

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．