证明sin(a b)=sinacosb cosasinb

(1+tana+cota)/(1+tan^2a+tana)-cota/(1+tan^2a)=(1+sina/cosa+cosa/sina)/(1+sina^2/cosa^2a+sina/cosa)-c

首先sin(α)=sin((α+β)-β)=sin(α+β)cos(β)-cos(α+β)sin(β)cos(α)=cos((α+β)-β)=cos(α+β)cos(β)+sin(α+β)sin(β)

(tanα∙sinα)/(tanα-sinα)=(tanα∙sinα)/[sinα(1/cosα-1)]=tanα/[(1/cosα)-1)]=(tanα∙cosα

由和差化积公式知：sinx+sinθ=2sin(x+θ)/2cos(x-θ)/2sinx-sinθ=2sin(x-θ)/2cos(x+θ)/2故(sinX+sinθ).(Sinx-sinθ)=2sin

证明：如图所示作单位圆,设∠AOC=α,∠COD=β,则∠AOD=α+β,AO=1作AB⊥Ox交Ox于B,作AC⊥OC交OC于C,作CE⊥AB交AB于E,作CD⊥Ox交Ox于D易证△OBF∽△ACF∴

[sinα+cos(α+β)sinβ]*cos(α+β)=sinαcos(α+β)+(1-(sin(α+β))^2)sinβ=(1/2)(sin(2α+β)-sinβ)+sinβ-(sin(α+β))

sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2（sinx）^2-1+2(siny）^2)=（si

sinπ(x-1)=sin(πx-π)=-sinπx

左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin²a(1-sin

因为sin(x)与arcsin(x)互为反函数,根据反函数的性质f[f-1(x)]=x可得sin(arcsinx)=x

奇变偶不变,符号看象限.比如（1）中,360°等于180°乘2,2是偶数,所以函数名不变,后面先写上sinα,然后把α看成锐角,-α就是第四象限角,加360°后仍是第四象限角,第四象限角的sinα（看

sinπ(x-1)=sin(πx-π)=sinπxcosπ-cosπxsinπ=-sinπx-0=-sinπx再问：那个。。。从第二步怎么化成第三步再答：不是有公式sin(a-b)=sinacosb-

(1)sin(A＋B)/sin(A-B)=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)=3,(tanA+tanB)/(tanA-tanB)=3,tanA+tanB=3

左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin

应用数学归纳法.1.当n=1时,左边=sin(pi/3),右边=sin(pi/3).则命题成立2.假设当n=k时,命题成立.即sin(pi/3)+sin(2*pi/3)+...+sin(k*pi/3)

tanαsinα/(tanα-sinα)=sinα/cosα*sinα/(sinα/cosα-sinα)=[(sinα)^2/cosα]/[(sinα-sinαcosα)/cosα]=[(sinα)^

(tan^2-sin^2)cot^2=(sin^2/cos^2-sin^2)cot^2=sin^2(1/cos^2-1)cot^2=sin^2(1-cos^2)/cos^2*cot^2=[sin^2*

sin(A+B)=sinAcosB+cosAsinB证明如图我们先来证明cos(A-B)=cosAcosB+sinAsinB在标准圆中.AB为直径.长度为1  由圆的性质可知角AD

sin(360°-a）=sin(-a)=-sina

应该是sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]A=(A+B)/2+(A-B)/2.B=(A+B)/2-(A-B)/2所以sin(A+B)/2cos(A-B)/2+cos(