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 = 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 !!
To express 3 cos(x) - 2 sin(x) in the form R cos(x + a), we can make use of the trigonometric identity:
R cos(x + a) = R cos(x) cos(a) - R sin(x) sin(a)
Comparing this with the given expression, we equate the coefficients:
3 cos(x) = R cos(a)
-2 sin(x) = -R sin(a)
Dividing the two equations, we get:
(3 cos(x))/(-2 sin(x)) = (R cos(a))/(-R sin(a))
Simplifying further, we have:
-3/2 = cot(a)
To find the values of a, we need to determine the arccot(-3/2). Using a calculator, we find that:
a ≈ 151.4°
Now, substituting this value of a back into the equation R cos(x) = 3 cos(a), we can solve for R:
R cos(x) = 3 cos(a)
R cos(x) = 3 cos(151.4°)
R cos(x) = -1.5
Therefore, the expression 3 cos(x) - 2 sin(x) can be written as -1.5 cos(x + 151.4°).
To find the maximum and minimum values of this expression, we can use the fact that the maximum value of cos(x + a) is 1, and the minimum value is -1. Therefore, the maximum value of -1.5 cos(x + 151.4°) is -1.5 and the minimum value is 1.5.
Hence, the maximum value of 3 cos(x) - 2 sin(x) is -1.5 and the minimum value is 1.5.