Trig Identities

Shades.......we have to be careful about multiplying across these identities like the guest did......we might end up with something that isn't true  = not an identity.....here's another (better) approach

(2 + cos2y - 5 sin y)/(3 + sin y) = 1 - 2 sin y        [ remember cos2y = 1 - 2sin^2y]

(2 + 1 - 2sin^2y - 5siny) / (3 + siny)      simplify the numerator

(-2sin^2y - 5siny + 3) / ( 3 + siny)     factor the numerator

(siny + 3)(-2siny + 1) / (3 + siny)     rearrange the factors slightly

( 3 + siny) (1 -2siny) / (3 + siny)            3 + siny  "cancels" on top/bottom

1 - 2siny    =   1 - 2siny

Hello ? ?

Stop trolling!!!!!!!!!!!!!!!!!!!

Verify the following identity:

(2+cos(2 y)-5 sin(y))/(3+sin(y)) = 1-2 sin(y)

Multiply both sides by 3+sin(y):

2+cos(2 y)-5 sin(y) = ^?(1-2 sin(y)) (3+sin(y))

cos(2 y) = 1-2 sin(y)^2:

2+1-2 sin(y)^2-5 sin(y) = ^?(1-2 sin(y)) (3+sin(y))

2+1-2 sin(y)^2-5 sin(y) = 3-5 sin(y)-2 sin(y)^2:

3-5 sin(y)-2 sin(y)^2 = ^?(1-2 sin(y)) (3+sin(y))

(1-2 sin(y)) (3+sin(y)) = 3-5 sin(y)-2 sin(y)^2:

3-5 sin(y)-2 sin(y)^2 = ^?3-5 sin(y)-2 sin(y)^2

The left hand side and right hand side are identical:

Answer: |(identity has been verified)