Find the Minimum Value of 4 cos x − 3 sin x + 7

Solution

For an expression of the form

\[ a\cos x+b\sin x \]

the maximum value is

\[ \sqrt{a^2+b^2} \]

and the minimum value is

\[ -\sqrt{a^2+b^2} \]

Here,

\[ a=4, \qquad b=-3 \]

Therefore,

\[ \sqrt{a^2+b^2} = \sqrt{4^2+(-3)^2} \]

\[ = \sqrt{16+9} \]

\[ = \sqrt{25} =5 \]

Hence,

\[ -5 \leq 4\cos x-3\sin x \leq 5 \]

Adding \(7\) throughout,

\[ 2 \leq 4\cos x-3\sin x+7 \leq 12 \]

Therefore, the minimum value is

\[ \boxed{2} \]