作业帮sin瞲 cos硏 dx 0 to Π
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=（cos15°）平方-（sin15°）平方=cos（15+15）=（根号3）/2
原式=根号2*(根号2/2*cosπ/12—根号2/2*sinπ/12)=根号2*(sinπ/4cosπ/12—cosπ/4sinπ/12)=根号2*sin(π/4-π/12)=根号2*sin(π/6
＝（1/2）sinπ/6＝（1/2）×（1/2）＝1/4这题是代入sin2α＝2sinαcosα求得的
0sin(cosx),当pi/4
sin(π+a)cos(-a)+sin(2π-a)cos(π-a)=-sinacosa+sinacosa=0
[sin(π/12)+cos(π/12)]^2=[sin(π/12)]^2+2sin(π/12)*cos(π/12)+[cos(π/12)]^2=1+22sin(π/12)*cos(π/12)=1+s
即sin90度等于一；cos90度等于零,这个很好记的sin即九十度y轴的值；cos即x轴
√2sin(x/2+π/4)=sinx/2+cosx/2tan(x/2-π/4)=-cot(x/2+π/4)a*b=√2sin(x+π/4)f(x)的导数=√2cos(x+π/4)f(x)加上f(x)
cosπ/4cosπ/3-sinπ/4sinπ/3=cos45°cos60°-sin45°sin60°=(√2/2)*(1/2)-(√2/2)*(√3/2)=(√2-√6)/4
sin(πcosx)=cos(πsinx)πcosx+πsinx=π/2cosx+sinx=1/2平方1+2sinxcosx=1/4sinxcosx=-3/8所以sinx=(1±√7)/4x=arcs
[cos(a-π)/sin(π-a)]sin(a-π/2)cos(π/2+a)=[-cosa/sina](-cosa)(-sina)=-cos^2a再问：还有一个，在下面~再答：化简[cos(2π-a
sin(2π-a)=-sinacos(π/2-a)=sinacos(π-a)=-cosasin(π/2+a)=cosasin(2π-a)cos(π/2-a)/cos(π-a)sin(π/2+a)=ta
原式=2sin(π/10)cos(π/10)cos(π/5)/2cos(π/10)=sin(π/5)cos(π/5)/2cos(π/10)=2sin(π/5)cos(π/5)/4cos(π/10)=s
-sinacosasina=-sin^2acosa
是要化简吗?(cosπ/8+sinπ/8)(cos³3π/8-sin³3π/8)=(cosπ/8+sinπ/8)(cosπ/8-sinπ/8)(cos²3π/8+cos3
先证明f（x）=sinx-xx∈（0,π/2）是一个减函数,可知sinx
θ∈(0,π/2)sinθcosθθ∈(0,π/2)cosθ∈(0,1)又sinθ
1sin²α+cosαcos(π/3+α)-sin²(π/6-α)=sin²α+cosα(cosπ/3cosα-sinπ/3sina)-(sinπ/6cosα-cosπ/
(cosπ/8+sinπ/8)(cos^3π/8+sin^3π/8)=[(cosπ/8+sinπ/8)^2](cos^2π/8-cosπ/8sinπ/8+sin^2π/8)=(1+2cosπ/8sin