Basic Trig (r-sin) help!

**Fezzan** · September 4th 2010, 02:34 AM

Hey guys,

Sorry I couldn't have a more direct title. There isn't really a name for this exact topic, but anyway, this is a rule we learnt:

asinθ + bcosθ = r sin (θ + λ)

where r = √a^2 + b^2
and tanλ = b/a

Question: Solve for 0°≤ θ ≤ 360°:
√2Cosθ + Sinθ = -1

My Working:

r = √3
λ = 54 degrees 44 minutes.

r sin (θ + λ) = -1
√3sin(θ + 54°44') = -1
sin(θ + 54°44') = -1/√3
θ + 54°44' = ?

I know you minus 54°44' from the answers you get. But from here onwards, I do not understand what to do. I've tried with answers but I'm failing.

Answers are 160°32', 270° for 0°≤ θ ≤ 360°.

Please help!
Many thanks.

**Prove It** · September 4th 2010, 04:13 AM

For your question

$\sqrt{2}\cos{\theta} + \sin{\theta} = -1$

$(\sqrt{2}\cos{\theta} + \sin{\theta})^2 = (-1)^2$

$2\cos^2{\theta} + 2\sqrt{2}\cos{\theta}\sin{\theta} + \sin^2{\theta} = 1$

$\cos^2{\theta} + 2\sqrt{2}\cos{\theta}\sin{\theta} + 1 = 1$ since $\cos^2{\theta} + \sin^2{\theta} = 1$

$\cos^2{\theta} + 2\sqrt{2}\cos{\theta}\sin{\theta} = 0$

$\cos{\theta}(\cos{\theta} + 2\sqrt{2}\sin{\theta}) = 0$

$\cos{\theta} = 0$ or $\cos{\theta} + 2\sqrt{2}\sin{\theta} = 0$ .

Case 1: $\cos{\theta} = 0$

$\theta = \frac{\pi}{2} + n\pi$ where $n \in \mathbf{Z}$ .

Case 2: $\cos{\theta} + 2\sqrt{2}\sin{\theta} = 0$

$\cos{\theta} = -2\sqrt{2}\sin{\theta}$

$\frac{\cos{\theta}}{\sin{\theta}} = -2\sqrt{2}$

$\frac{1}{\tan{\theta}} = -2\sqrt{2}$

$\tan{\theta} = -\frac{1}{2\sqrt{2}}$

$\tan{\theta} = -\frac{\sqrt{2}}{4}$

$\theta = n\pi - \arctan{\frac{\sqrt{2}}{4}}$ , where $n \in \mathbf{Z}$ .

**Fezzan** · September 4th 2010, 05:43 AM

Thanks very much for that detailed answer...

Although, I probably should've said this in the my first post: I'm studying for an exam that requires me to specifically use the r sin (θ + λ) method. Could you just help continue from the working I already did in my first post? I'd very much appreciate your further help.

But thanks anyway for your answer... I'm trying to specifically find angles for the equation though... i.e. 160°32' & 270°. I'm only in Grade 11 maths... I didn't really understand your answer :P

**Opalg** · September 4th 2010, 06:06 AM

Originally Posted by Fezzan

Question: [/I]Solve for 0°≤ θ ≤ 360°:
√2Cosθ + Sinθ = -1

My Working:

r = √3
λ = 54 degrees 44 minutes.

r sin (θ + λ) = -1
√3sin(θ + 54°44') = -1
sin(θ + 54°44') = -1/√3
θ + 54°44' = ?

Your solution is completely correct up to that point. To complete the question, you need to find angles whose sine is –1/√3. There are various ways to do that. One method is to find the angles whose sine is +1/√3, namely 35°16' and 180° – 35°16' = 144°44'. Now use the fact that sin(x+180°) = –sin x. This tells you that the angles whose sine is –1/√3 are 35°16' + 180° = 215°16' and 144°44' + 180° = 324°44'. Finally, subtract 54°44' from those angles to get your answers.

If you're looking for extra credit, you might like to think about why one of the answers turns out to be exactly 270° (three right angles).

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