trig

Daniel123 · 2008-02-02 07:08:15

Hello.

Can someone tell me what I'm doing wrong here?

Solve in the interval [0,90°]: 2cos3θ - 3sin3θ = -1

My working:

Let 2cos3θ - 3sin3θ ≡ Rsin(3θ - α)
2cos3θ - 3sin3θ ≡ Rsin3θcosα - Rcos3θsinα
∴ Rsinα = -2, Rcosα = -3
⇒ tanα = 2/3, α ≈ 33.7

R = √(3² + 2²) = √13

∴ √13sin(3θ - 33.7) = -1
sin(3θ - 33.7) = - √13/13

Then I used the quadrant diagram, with an acute angle of 16.1 in the 3rd and 4th quadrants.

⇒ 3θ - 33.7 = 196.1
θ = 76.6

This answer is wrong, and I can't see why? Can anyone help?

Thanks.

JaneFairfax · 2008-02-02 07:45:48

You had both Rsin͍α < 0 and Rcosα < 0; therefore your α is in the third quadrant. So α = 213.7°.

Daniel123 · 2008-02-02 08:28:45

Ahh of course... thanks again Jane.

Daniel123 · 2008-02-02 09:01:19

I still get the wrong answer.

3θ - 213.7 = 196.1
θ = 136.6, which is out of range, so θ = 180 - 136.6 = 43.4

The 136.6 works, but 43.4 doesn't. :s

Last edited by Daniel123 (2008-02-02 09:06:55)

JaneFairfax · 2008-02-02 09:16:43

That should give

Last edited by JaneFairfax (2008-02-02 09:20:14)

Daniel123 · 2008-02-02 09:29:39

Silly me. Thanks .

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