Trigonometry problems help!!!

If x=3sinθ and y=3cosθ - 2, elimination θ to find an equation relating x and y.

The answer is x^2+y^2+4y-5=0

THANKS!

sin(theta) = x/3
cos(theta) = (y + 2)/3

remember that sin^2 (theta) + cos^2 (theta) = 1

(x / 3) ^2 + [(y + 2)/3]^2 = 1

x^2 /9 + [ (y^2 + 4y + 4) / 9 ] = 1 Multiply through by 9
x^2 + y^2 + 4y + 4 = 9
x^2 + y^2 + 4y - 6 = 0

Last line contains a typo.
Sorry.
It really should be - 5 not - 6

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Try to think of identities that relate sine and cosine. Two are sin(π/2- θ) = cosθ and
sin^2 θ + cos^2 θ = 1
The pythagorean identity is better because you can eliminate both sine and cosine at the same time.
So square both equations to get the sin θ and cos θ to look like the identity.
x^2 = 9 sin^2 θ
The second equation comes out y^2 = 9cos^2 θ -12cos θ +4. This is not what you want because there is an extra cos θ term. So to avoid that, add the two to both sides then square both sides to get:
(y+2)^2 = (3cos θ)^2
y^2 + 4y + 4 = 9cos^2 θ
Now add both equations together since that is what's happening in the identity
x^2 + y^2 + 4y + 4 = 9 sin^2 θ + 9cos^2 θ
Next eliminate the θ by factoring out the 9 and using the identity
x^2 + y^2 + 4y + 4 = 9 (sin^2 θ + cos^2 θ)
x^2 + y^2 + 4y + 4 = 9 (1)
x^2 + y^2 + 4y + 4 - 9 = 9 - 9
x^2 + y^2 + 4y - 5 = 0

x = 3*sin(θ)

y = 3*cos(θ) - 2

Recall that x = r*cos(θ), y = r*sin(θ), and x² + y² = r²

x = 3*sin(θ) --> (i)

y + 2 = 3*cos(θ) --> (ii)

(i)² + (ii)² = x² + (y + 2)² = 9*sin²(θ) + 9*cos²(θ)

x² + y² + 4y + 4 = 9

x² + y² + 4y - 5 = 0 as desired

x = 3 sin(θ) → cos(θ) = √ ( 1 - (x/3)² )

... y = 3 cos(θ) - 2
or y = 3 √ ( 1 - (x/3)² ) - 2
or y = √ ( 9 - x² ) - 2
or ( y + 2 ) = √ ( 9 - x² )
or y² + 4y + 4 = 9 - x²
or x² + y² + 4y - 5 = 0 ← you're absolutely correct - bravo !