maths - trigo2 (proof)

Typing error ??

i) sin(A+B) sin(A-B) should be - sin(A+B) sin(A-B) ??

ii) b) cos²A - cos²B + cos²C = - 1
should be cos²A - cos²B + cos²C = 1 ??

2011-07-27 23:51:06 補充：
i)cos²A - cos²B= (cosA - cosB) (cosA + cosB)= [ - 2sin (A+B)/2 sin(A-B)/2 ] [ 2cos (A+B)/2 cos(A-B)/2 ]= - 2 sin (A+B)/2 cos (A+B)/2 * 2 sin(A-B)/2 cos(A-B)/2= - sin(A+B) * sin(A-B) = sin(A+B) sin(B-A)
ii)a)cos²A - cos²B + sin²C= sin²C - sin(A+B) sin(A-B)= sin²C - sin(π - C) sin(A-B)= sin²C - sinC sin(A-B)= sinC (sinC - sin(A-B))= sinC (sin(A+B) - sin(A-B))= sinC ( sinAcosB + cosAsinB - (sinAcosB - cosAsinB) )= sinC (2 cosAsinB )= 2 cosA sinB sinCb)cos²A - cos²B - cos²C = - 1cos²A - cos²B + (1 - cos²C) = 0cos²A - cos²B + sin²C = 0By the result of a) :2 cosA sinB sinC = 0cosA = 0 or sinB = 0 (rejected since sinB≠0 for 0 < B < π) or sinC = 0(rejected)A = π/2Therefore △ABC is a right-angled △.
iii)cos²x - sin²y= cos²x - cos²(π/2 - y)By the result of i) ,= - sin (x + π/2 - y) sin (x - π/2 + y)= sin (π/2 + x - y) sin (π/2 - (x + y))= cos -(x - y) cos(x + y)= cos (x + y) cos (x - y)


2011-07-28 13:34:33 補充：
Yes! No need to write cos -(x - y) cos(x + y).

but cos -(x-y) = cos(x-y) is no problem.