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#1 2007-10-23 08:58:02

Daniel123
Member
Registered: 2007-05-23
Posts: 663

more trig

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 smile

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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#2 2007-10-23 09:52:09

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: more trig

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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#3 2007-10-23 10:03:52

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: more trig

Thank you smile

mathsyperson wrote:

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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