Hi

Forgive me if this is the incorrect forum - I consider compound angles to be trig, but maybe they're not!

I'm struggling to obtain the same answer as that given in a textbook. I'll give you my answer - I'd appreciate any help where I might have gone wrong:

Convert $\displaystyle -3 \, sin\,\omega t-6 \,cos\,\omega t $ to the form $\displaystyle R\,sin\,(\omega t\pm \alpha) $

I make this $\displaystyle 6.708sin(\omega t- 4.2487)$

Is this in the 3rd quadrant as I think it is? The answer the book gives seems to imply the answer is in the second quadrant.