Find the Minimum Value of 3 cos x + 4 sin x + 8

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,

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

Therefore,

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

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

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

Hence,

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

Adding \(8\) throughout,

\[ 3 \leq 3\cos x+4\sin x+8 \leq 13 \]

Therefore, the minimum value is

\[ \boxed{3} \]

Final Answer

\[ \boxed{ \text{Minimum value}=3 } \]

Correct Option: (d)