Discussion on Trigonometric Ratios and Identities

Prove sin2[square]x+cos2x=1 sin0=0

sin(x)^2+cos(x)^2 = 1

cos(x)^2+sin(x)^2 = 1

cos(x)cos(x)+sin(x)sin(x) = 1

cos(x - x) = 1 ....

Note: I'm using the Sum-Difference Identity: cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

cos( 0 ) = 1

1 = 1

 do it by sin x= perp / hyp ,etc 

As perp^2 + Base ^2= Hyp^2 (Pythagoras), it is 1.