For your question
, where .
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
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°.
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
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).