trigonometry
posted by lindsay .
express 3 cos x 2 sin x in th eform R cos (x + a) and hence write down the maximum and minimum values of 3 cos x  2 sin x.

let 3cosx  2sinx = Rcos(x+a)
Rcos(x+a) = R(cosxcosa  sinxsina)
= Rcosxcosa  Rsinxsina
so we have the identity
Rcosxcosa  Rsinxsina = 3cosx2sinx
this must be valid for any x
so let's pick x's that simplify this
let x = 0
then
Rcos0cosa  Rsin0sins = 3cos0  2sin0
Rcosa = 3
cosa = 3/R
let x = 90°
Rcos90cosa  Rsin90sina = 3cos90  2sin90
Rsina = 2
sina = 2/R
but sin^2a + cos^2a = 1
4/R^2 + 9/R^2 = 1
R^2 = 13
R = √13
also : sina/cosa = (2/R) / (3/R) = 23
tana = 2/3
a = arctan (2/3) = 33.69°
thus 3cosx  2sinx = √13cos(x + 33.69°)
check by taking any angle x
let x = 26°
LS =1.8196...
RS = √13 cos(5969) = 1.8196
........ how about that !!