trig 3

[red and white kop!]

express cosA - (sqrt3 x sinA) in the form r x sin (A-B)
finding r and B, obviously
so using r= sqrt (a^2 + b^2) i found r=2, which is wrong as the answer is -2
i know that this must be correct as i drew the diagram and the sqrt allows for it, but how is the sign exactly defined and why isnt this question asked in every case of defining r? there was no mention of the ambiguity of the sqrt in my lesson.

 

[Deleted member 4993]

express cosA - (sqrt3 x sinA) in the form r x sin (A-B) <<< Do not write "x" for multiplication - Use " * " insteadfinding r and B, obviously
so using r= sqrt (a^2 + b^2) i found r=2, which is wrong as the answer is -2
i know that this must be correct as i drew the diagram and the sqrt allows for it, but how is the sign exactly defined and why isnt this question asked in every case of defining r? there was no mention of the ambiguity of the sqrt in my lesson.

\(\displaystyle 1\cdot cos(A) - \sqrt{3} \cdot sin(A) = 2 \cdot sin (\frac{\pi}{6} - A) = -2 \cdot sin (A - \frac{\pi}{6})\)

Where is the ambiguity?!