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.

Answer 1

let 3cosx - 2sinx = Rcos(x+a)

Rcos(x+a) = R(cosxcosa - sinxsina)
= Rcosxcosa - Rsinxsina

so we have the identity
Rcosxcosa - Rsinxsina = 3cosx-2sinx
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 !!