Prove the identity (cosx+cosy)^2+(sinx-siny)^2= 2+2cos(x+y) if at all possible, provide a two-columned answer, explaining each step. you : d

(cos(x) + cos(y))^2 + (sin(x) - sin(y))^2 Remove the brackets

cos^2(x) + cos^2(y) + 2cos(x)*cos(y) + sin^2(x) - 2(sin(x)*sin(y) + sin^2(y) Combine these two in bold to make 1 because sin^2(x) + cos^2(x) = 1

1 + cos^2(y) + 2cos(x)*cos(y) - 2*sin(x)*cos(y) + sin^2(y) 
These two in bold also make 1

2 + 2cos(x)*cos(y) - 2*sin(X)*sin(y) Bring out a common factor of 2
2 +2(cos(x)*cos(y) -  sin(x)*sin(y) )

but cos(x+y ) = cos(x)*cos(y) - sin(x)*sin(y)

2 + 2* cos(x + y) is your final answer. 

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm