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I just want to check my answers for this as, yet again, I disagree with my textbook.
(a) Factorise 4xy - y² + 4x - y
(b) Solve the equation 4 sinθ cosθ - cos²θ + 4sinθ - cosθ = 0
Thanks
Edit: And another:
Express 4 cos3θ - sin (90 - 3θ) as a single trigonometric function (could you please show/tell me how you did this, as I don't know how to).
Last edited by Daniel123 (2007-10-23 09:16:53)
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a) 4xy - y² + 4x - y = (4x-y)(y+1).
b) 4 sinθ cosθ - cos²θ + 4sinθ - cosθ = 0
Using a, this becomes (4sinθ - cosθ)(cosθ + 1) = 0.
Therefore, either 4sinθ - cosθ = 0 or cosθ + 1 = 0.
For the first case:
4sinθ - cosθ = 0
4sinθ = cosθ
4tanθ = 1
tanθ = 1/4
θ ≈ 14.0°
For the second:
cosθ + 1 = 0
cosθ = -1
θ = 180°.
So, for 0°≤θ<360°, the solutions are θ = ~14.0°, 180° and ~194.0°.
The third question is simple if you know that sin θ ≡ cos (90-θ).
Why did the vector cross the road?
It wanted to be normal.
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Thank you
The third question is simple if you know that sin θ ≡ cos (90-θ).
I did not know that... but it's actually quite obvious. Woops.
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