Find the Maximum and Minimum Value of 12 sin x − 5 cos x

Solution

We use the standard result:

\[ a\sin x + b\cos x \] has maximum value \[ \sqrt{a^2+b^2} \] and minimum value \[ -\sqrt{a^2+b^2} \]

Given expression:

\[ 12\sin x – 5\cos x \]

Here, \[ a=12,\qquad b=-5 \]

Now,

\[ \sqrt{a^2+b^2} = \sqrt{12^2+(-5)^2} \]

\[ = \sqrt{144+25} \]

\[ = \sqrt{169} \]

\[ =13 \]

Therefore,

Maximum value: \[ 13 \]

Minimum value: \[ -13 \]

Final Answer

\[ \boxed{\text{Maximum value }=13} \]

\[ \boxed{\text{Minimum value }=-13} \]