Question 769258
<pre>
cos(u+v)cos(v) + sin(u+v)sin(v) = cos(u)

Remembering the identity

cos(A-B) = cos(A)cos(B) + sin(A)sin(B)

we recognize that the left side,

cos(u+v)cos(v) + sin(u+v)sin(v)

is simply cos(A)cos(B) + sin(A)sin(B)

with u+v substituted for A and v substituted for B.

 we have

cos[(u+v)-v]

cos(u+v-v)

cos(u)

Edwin</pre>