sin cos problem

the equation is true,below i first prove that cos(x-y) = cos(x)*cos(y)+sin(x)*sin(y)
in a coordinate plane ,draw a circle with centre on origin and with radius 1 unit
let P,Q be two pts on this circle, join OP,OQ where P is a pt with greater angle in coordinate plane
let the inclination of OP be x degree and OQ's be y degrees, then anglePOQ=x-y
obviously,P=(cosx,sinx) Q=(cosy,siny)
by cosine rule, consider triangle POQ, PQ=[1+1-2*1*1cos(x-y)]^1/2
by distant form. PQ=[(cosx-coxy)^2+(sinx-siny)^2]^1/2
So,1+1-2*1*1cos(x-y)=(cosx-coxy)^2+(sinx-siny)^2
2-2cos(x-y)=(cosx)^2+(cosy)^2+(sinx)^2+(siny)^2-2cosxcosy-2sinxsiny
2-2cos(x-y)=2-2cosxcosy-2sinxsiny
So, cos(x-y) = cos(x)*cos(y)+sin(x)*sin(y)
We simply put y=-y into cos(x-y) = cos(x)*cos(y)+sin(x)*sin(y)
We get cos(x+y) =cos(x)*cos(-y)+sin(x)*sin(-y)
That is cos(x+y) = cos(x)*cos(y)-sin(x)*sin(y)

2007-04-29 17:58:34 補充：
sorry,the word inclination should be changed to ANGLE,that is the angle measuresfrom the x- positive axis to the line OP and OQ

yes,let me proved it for you then=]
you know that sin(x-y)= sinXcosY-cosXsinY
we can replace X by (180/2)-X
then we will have:
sin( (180/2)-X-Y) = sin (180/2-X)cosY-cos(180/2-X)sinY
after simply
sin( (180/2)-X-Y) = cos(X+Y)
therefore
cos(X+Y)=cosXcosY-sinXsinY