sin+2arcsin+等于什么

贲雯全2974在三角形ABC中,设sin^2(A/2)+2sin^2(B/2)+sin^2(C/2)=1 -
咸览温13789375772 ______ 解:半角的正弦公式(降幂扩角公式) sin^2(α/2)=(1-cosα)/2 得:2sin^2(B/2)=1-cosB sin^2(A/2)+sin^2(C/2)=1-1/2(cosA+cosC)=cosB cosA+cosC=2(1-cosB)=4sin^2(B/2)……① 在△ABC中,A+B+C=π,于是 B/2=(π-A-C)/2,sinB/2=sin(π-A-C)/2=...

贲雯全2974sin2x+cos2x+2=根号2倍的sin(2x+π/4)+2怎么出来的,求详细过程 -
咸览温13789375772 ______ sin(2x)+cos(2x)+2=√2[(√2/2)sin(2x)+(√2/2)cos(2x)]+2=√2[sin(2x)cos(π/4)+cos(2x)sin(π/4)]+2=√2sin(2x+π/4)+2

贲雯全2974化简sin(α+2π)cos(π+α)/sin( - α - π)cos( - π - α) -
咸览温13789375772 ______ 原式=[sina*(-cosa)]/[-sin(a+π)cos(π+a)]=(-sinacosa)/(-sinacosa)=1

贲雯全2974已知sinβ+2sin(2α+β)=0且α≠kπ/2,α+β≠π/2+kπ(k∈Z) -
咸览温13789375772 ______ β=(α+β)-α,2α+β=(α+β)+α sinβ=sin(α+β)cosα-cos(α+β)sinα,sin(2α+β)=sin(α+β)cosα+cos(α+β)sinα 由sinβ+2sin(2α+β)=0得sin(α+β)cosα-cos(α+β)sinα+2sin(α+β)cosα+2cos(α+β)sinα=0,即3sin(α+β)cosα+cos(α+β)sinα=0,除以cos(α+β)cosα,得3tan(α+β)+tanα=0

贲雯全2974sin(arcsin1/2+arccos1/3)? -
咸览温13789375772 ______ sin(arcsin1/2+arccos1/3)=sin(arcsin1/2)cos(arccos1/3)+cos(arcsin1/2)sin(arccos1/3)=1/2*1/3+根号(1-(1/2)^2)*根号(1-(1/3)^2)=1/6+(根号3)/2*2根号(2)/3=1/6+(根号6)/3=(1+2根号6)/6

贲雯全2974(sin(ln2/2)+sin(ln3/3)+...+sin(lnn/n))^(1/n)的极限 -
咸览温13789375772 ______ sin(lnk/k)

贲雯全2974sin^2(a)+sinacosa+2 怎么化简啊 大佬 -
咸览温13789375772 ______ 数学大牛来帮你!(sina-sinp)2=1/4(cosa-cosp)2=1/9 (sina)^2+(sinp)^2-2sinasinp =1/4 ...1 (cosa)^2+(cosp)^2-2cosacosp=1/9 ...21+22-2sin...

贲雯全2974sinα/2+sinα的值域 -
咸览温13789375772 ______ f(x)=sinx/2+sinx=sinx/2(1+2cosx/2)=sinx/2[1+2(1-sin^2x/2)]=sinx/2(3-2sin^2x/2)=-2sin^3x/2+3sinx/2 令t=sinx/2,则有原式y=-2t^3+3t,t[-1.1],对其进行求导 则y'=-6t^2+3,故t=正二分之更号二或负二分之更号二时取得极值,并分析得出,当t[-1,负二分之更号二]为单调减函数,[负二分之根号二,正二分之更号二]为单调增,[正二分之更号二,1]为单调减,故可知 f(x)的值域为[负更号二,正更号二]

贲雯全2974sin(arcsin1/2+arccos1/3)? -
咸览温13789375772 ______[答案] sin(arcsin1/2+arccos1/3)=sin(arcsin1/2)cos(arccos1/3)+cos(arcsin1/2)sin(arccos1/3)=1/2*1/3+根号(1-(1/2)^2)*根号(1-(1/3)^2)=1/6+(根号3)/2*2根号(2)/3=1/6+(根号6)/3=(1+2根号6)/6...

贲雯全2974求证:三角函数sin2A+sin2B+sin2C=4sinAsinBsinC -
咸览温13789375772 ______ 证明:∵sin2A+sin2B+sin2C=sin2A+sin2B+sin2(π-A-B)=sin2A+sin2B+sin(2π-2A-2B)=sin2A+sin2B+sin(-2A-2B)=sin2A+sin2B-sin(2A+2B)=sin2A+sin2B-sin2Acos2B-cos2Asin2B=sin2A(1-cos2B)+sin2B(1-cos2A)=2sin2Asin²B+sin2Bsin²A...