100√3sin

35°再问：分数表示下，谢谢

sin³x-sin^5x=sin³x(1-sin²x)=sin³xcos²x当00√(sin³xcos²x)=sinxcosx√s

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间：π/2+2

√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)

f(x)=2sin(2wx-π/6)+λw=1/3+k/2(k=0,+_1.)w=5/6f(x)=2sin(5/3x-π/6)+λ(1)T=6π/5(2)f(x)=2sin(5/3x-π/6)-根号2

令³√x=t,那么x=t^3,用分部积分法来慢慢做,但要细心原积分=∫sintd(t^3)=∫3t^2*sintdt=-3t^2*cost+∫cost*d(3t^2)=-3t^2*cost+

我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

先利用正弦定理将角都转为边再用余弦定理即b^2-a^2-c^2=√3acb^2=a^2+c^2+√3ac=a^2+c^2-2accosBcosB=-√3/2又B∈(0,π)B=120°

sin(θ/2+3π/2)>0-cos(θ/2)>0cos(θ/2)

(1)f（x）=mn=(cosα+sinα)(cosα-sinα)+2√3sinαcosα=cos2α+√3sin2α=2[sin2α·cosπ/6+cos2α·sinπ/6]=2sin（2α+π/6

分子√1+2sinα(5π-α)*cos(α-π)是不是有问题?√1+2sin(5π-α)*cos(α-π)吧显然sin(5π-α)=sin(π-α)=sinα,而cos(α-π)=-cosα,故√1

f(x)=(√3/2)sin2x-(1/2)[(cosx)^2-(sinx)^2]-1=(√3/2)sin2x-(1/2)cos2x-1=sin(2x-π/6)-1f(x)的最大值是0,最小值是-2,

1+2sin(3π-a)cos(a-3π)=1+2sin(π-a)cos(a-π)=1-2sinacosa=(sina)^2-2sinacosa+(cosa)^2=(sina-cosa)^2a为第二象

f(x)=2cos(x-π/6)sin(x+π/6)-√3*(sin(x-π/6))^2+sin(x-π/6)cos(x-π/6）=sin2x+sin(π/3)-(√3/2）[1-cos(2x-π/3

设f(x)=100[sinθcosθ-(√3/3)sin²θ]=100[1/2sin2θ+(√3/6)(1-2sin²θ-1)]=100[1/2sin2θ+(√3/6)cos2θ-

/>利用积化和差公式,达到裂项的效果.2sinka*sin(a/2)=-cos[(k+1/2)a]+[cos(k-1/2)a]∴2sin(a/2)*(sina+sin2a+sin3a+...+sinn

sin(x-π/3)-cos(x+π/6)+√3cosx=sinxcosπ/3-cosxsinπ/3-cosxcosπ/6+sinxsinπ/6+√3cosx=1/2*sinx-√3/2*cosx-√

2（√3/2sinx+1/2cosx)=2sin(x+π/6）√2*√2（√2/2sinx-√2/2cosx)=2sin(x-π/4）（3）解;2√2(1/2cosx-√3/2sinx)=2√2cos

∵sin（π3-α）=sin[π2-（π6+α）]=cos（π6+α）=14，∴cos（π3+2α）=cos2（π6+α）=2cos2（π6+α）-1=2×（14）2-1=-78．故答案为：-78