sin(a+b)证明

怎样证明/sina+sinb/设f(x)=sin(x),它的导数是f'(x)=cos(x).利用微分中值定理,也就是拉格朗日定理,存在处于a与-b之间的实数ξ,使得f(a)-f(-b)=f'(ξ)[a-(-b)]即sina+sinb=cos

【证明】SinA+sinB=2Sin22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(

证明sin(a+b)sin(a-b)=sin^2a-sin^2b,左边=(sinacosb+cosasinb)(sinacosb-cosasinb)=sin²acos²b-cos²asin²b=sin

如何证明sin(a+b)sin(a-b)=sin*a-sin*b(*为平方)sin(a+b)sin(a-b)=(sinacosb+sinbcosa)(sinacosb-sinbcosa)==(sinacosb)^2-(sinbcosa)^2

证明sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2证明sin^2A+sin^2B-sin^2A*sin^2B+cos^2A*cos^2B=1Sin^2A+Sin^2B-Sin^2ASin^2B+Cos^2A

高中三角函数的证明,sin(A+B)*sin(A-B)=(sinA)^2-(sinB)^2但还不知道怎么证明.只要写下证明的思路和大致的过程就可以了说明下,sin(A+B)和sin(A-B)之间是乘号.式子的右边是sinA的平方减去sinB

sin(a+b)怎么证明的?我们老师说用向量证明但我觉得可以用三角形：把α、β两个角拼在一起.顶点为O,过B点作OB的垂线,交α另一边于A,交β另一边于C.则有S△OAC=S△OAB+S△OBC根据三角形面积公式,有sin(α+β)*OA*

证明sin(a+b)=sinacosb+cosasinb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△O

SIN(A-B)怎么证明的?

证明sin(A+B)=sinAcosB+cosAsinBsin(A+B)=cos[90-(A+B)]=cos[(90-A)-B]=cos(90-A)cosB+sin(90-A)sinB=sinAcosB+cosAsinB

证明sinasin(a+2b)-sinbsin(b+2a)=sin(a+b)sin(a-b)证明：sinasin(a+2b)-sinbsin(b+2a)=sin(a+b)sin(a-b)sinasin(a+b+b)-sinbsin(a+b+

如何证明sin(a+b)=sin(a)cos(b)+cos(α)sin(b)首先建立直角坐标系,在直角坐标系xOy中作单位圆O,并作出角a,b,与-b,使角a的开边为Ox,交圆O于点P1,终边交圆O于点P2,角b的始边为OP2,终边交圆O于

用正弦定理证明a+b/c=sinA+sinB/sinc应该是（a+b）/c=（sinA+sinB）/sinc吧?因为a=2RsinA,b=2RsinB,c=2RsinC,代入即可

证明sin(a+b)sin(a-b)=sin²a-sin²b,并用该式计算sin²20°+sin80°sin40°sin(a+b)sin(a-b)=(sinacosb+sinbcosa)(sinacosb-si

证明sin(A+B)cos(A-B)=sinAcosA+sinBcosBsin(a+b)cos(a-b)=(sinacosb+cosasinb)(cosacosb+sinasinb)=sinacosa(cosb)^2+(cosa)^2sin

懂得进.证明,sina+sin(a+b)+sin(a+2b)+...sin(a+nb)=sin(a+ab/2)sin[(n+1)b/2]/sin(b/2)数学归纳法或者[sina+sin(a+b)+sin(a+2b)+..+sin(a+nb

证明等式恒成立sin^a+sin^b-sin^asin^b+cos^acos^b=1sin^a+sin^b-sin^asin^b+cos^acos^b=1只需证sin^a(1-sin^b)+sin^b+cos^acos^b=1只需证sin^

证明sin(A+B)sin(A-B)=cos^2B-cos^2Asin(a+b)sin(a-b)=（sinacosb+cosasinb)(sinacosb-cosasinb)=sin^2acos^2b-cos^2asin^2b=(1-cos

a（sinB－sinC）＋b（sinC＋sinA）＋c（sinA－sinB）=0请证明...不好意思.我写错了.a（sinB－sinC）＋b（sinC—sinA）＋c（sinA－sinB）=0写错题目了吧?应该是：a（sinB－sinC）＋

证明三角函数等式sin(A+B)-sinA=2cos(A+B/2)sin(B/2)sin(A+B)-sinA=2cos(A+B/2)sin(B/2)和差化积sin(A+B)-sinA=1/2[cos(A+B)/2]*[sin(A+B-A)/