幾道比較難的數(shù)學(xué)··一時想不出望大家一起幫忙，謝謝～題目會用小號貼圖出來··

熱心網(wǎng)友

1。證:左邊=[cos(a+b)]^2-cos(a+b)cosb+[sin(a+b)]^2-sin(a+b)sinb=1-[cos(a+b)cosb+sin(a+b)sinb]=1-cos[(a+b)-b]=1-cosa=2[sin(a/2)]^2=右邊。2。(cosx+sinx)(cosx-sinx)(1+tg2xtgx)=[(cosx)^2-(sinx)^2]*[1+sin2xsinx/(cos2xcosx)]=cos2x*(cos2xcosx+sin2xsinx)/(cos2xcosx)=cos(2x-x)/cosx=cosx/cosx=13。因為cos2x=1/3且2x在第四象限。所以sin2x=-2√2/3,tanx=(1-co2x)/sin2x=3-2√2。原式=√3cos(p/4-x)/[1-sinx-(1-cosx)]=√3(√2/2*cosx+√2/2*sinx)/(cosx-sinx)=√6/2*(cosx+sinx)/(cosx-sinx)=√6/2*(1+tgx)/(1-tgx)=√6/2*(1-2√2)/(1+2√2)=√6/2*(1-2√2)^2/(-7)=-√6/14*(9-4√2)=(8√3-9√6)/14。

熱心網(wǎng)友

（1）左邊=(cosacosβ-sinasinβ)(cosacosβ-sinasinβ-cosβ)+(sinacosβ+cosasinβ)(sinacosβ+cosasinβ-sinβ)=cos^acos^β-cosacosβsinasinβ-cosacos^β-sinasinβcosacosβ+sin^asin^β+sinasinβcosβ+sin^acos^β+sinasinβcosacosβ-sinasinβcosβ+cosasinβsinacosβ+cos^asin^β-cosasin^β=cos^acos^β-cosacos^β+sin^asin^β+sin^acos^β+cos^asin^β-cosasin^β=cos^a(cos^β+sin^β)-cosa(cos^β+sin^β)+sin^a(sin^β+cos^β)=cos^a+sin^a-cosa=1-cosa右=2*[(1-cosa)/2]=1-cosa即證（2）（sinθ+cosθ）(cosθ+sinθ)(1+tg2θtgθ)=(cos^θ-sin^θ){1+(sin2θ/cos2θ)*[(1-cos2θ)/sin2θ]}=cos2θ*[1+(1-cos2θ)/cos2θ]=cos2θ*(1/cos2θ)=1(3)sin2x=-2√2/3,tanx=(1-co2x)/sin2x=3-2√2。原式=√3cos(p/4-x)/[1-sinx-(1-cosx)]=√3(√2/2*cosx+√2/2*sinx)/(cosx-sinx)=√6/2*(cosx+sinx)/(cosx-sinx)=√6/2*(1+tgx)/(1-tgx)=√6/2*(1-2√2)/(1+2√2)=√6/2*(1-2√2)^2/(-7)=-√6/14*(9-4√2)=(8√3-9√6)/14。

熱心網(wǎng)友

題目：如下