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sin16x化简

解2sin(16x)cos(8x) =sin24x+sin8x (应用的是积化和差公式:sinαcosβ = [sin(α+β)+sin(α-β)]/2)

都乘以sinx 原式=sinx*cosx*cos2x*cos4x*cos8x/sinx =1/2sin2x*cos2x*cos4x*cos8x/sinx =1/4sin4x*cos4x*cos8x/sinx =1/8sin8x*cos8x/sinx =1/16 sin16x/sinx 后面就很简单了吧.

sin^8x+cos^8x= =(sin^4x-cos^4x)^2+2sin^4xcos^4x =[(sin^2x-cos^2x)(sin^2x+cos^2x)]^2+2sin^4Xcos^4x =(sin^2x-cos^2x)^2+2sin^4Xcos^4x =sin^4x-2sin^2Xcos^2x+cos^4x+2sin^4Xcos^4x=sin^4x+cos^4x+2sin^4Xcos^4x-2sin^2Xcos^2x=(sin^4x+

-sinX

解:sinx+siny=√2 两边同时平方得:sinx+2sinxsiny+siny=2 ① cosx+cosy=√3/2 两边同时平方得:cosx+2cosxcosy+cosy=3/4 ② 由①+②得:2+2(sinxsiny+cosxcosy)=2+3/42(sinxsiny+cosxcosy)=3/4 2cos(x-y)=3/4 cos(x-y)=3/8

sin α+sin β-sinαsinβ+cosαcosβ =sin α+sin β-sinα(1-cosβ)+cosαcosβ =sin α+sin β-sinα+sinαcosβ+cosαcosβ =sin α+sin β-sinα+cosβ =1

解: 由于:sin^3(x)sin3x =sin^2(x)[sinx*sin3x] =sin^2(x)[sinx*sin(2x+x)] =sin^2(x)[sinx(sin2xcosx+sinxcos2x)] =sin^2(x)[sinxcosx*sin2x+sin^2(x)cos2x] =(1-cos2x)/2*[(1/2)(2sinxcosx)sin2x+(1-cos2x)/2*cos2x] =[(1-cos2x)/4]*[sin^2(2x)+(1-cos2x)cos2x]

这样的 原式=1-(cosa)-(sina) =1-(1-sina)-(sina) 令sina=t 所以原式=1-(1-t)-(t) =1-(1-3t+3t^4-t^6)-t^6 =1-1+3t-3t^4+t^6-t^6 =3t-3t^4 =3t(1-t) =3sina(1-sina) =3sinacosa =3/4*(2sinacosa) =(3/4) *sin2a

((1+cos20°)/2sin20°)-sin10°(cot5°-tan5°)=((1+cos20°)/4sin10°cos10°)-sin10°(cot5°-tan5°)=(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)=(cos10°/2sin10°)-2((cos5°)^2-(sin5°)^2)=(cos10°/2sin10°)-2cos10°=(cos10°-4sin10°cos10°)/2sin10°=(

sin(arctanT)与cos(arctanT)的化简 【解】令θ=arctanT,其中θ∈(-π/2,π/2)【非常重要】 那么tanθ=T,T=sinθ/cosθ 1+T=1+(sinθ/cosθ)=(sinθ+cosθ)/cosθ=1/cosθ 同理,1+(1/T)=1/sinθ 所以cosθ=1/(1+T),sinθ=T/(1+T) θ∈(

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