Find the Maximum and Minimum Value of 5 cos x + 3 sin(π/6 − x) + 4
Find the maximum and minimum values of the following trigonometrical expression: \[ 5\cos x + 3\sin\left(\frac{\pi}{6}-x\right)+4 \]
Solution
Using the identity:
\[ \sin(A-B)=\sin A\cos B-\cos A\sin B \]
We get,
\[ \sin\left(\frac{\pi}{6}-x\right) = \sin\frac{\pi}{6}\cos x-\cos\frac{\pi}{6}\sin x \]
\[ = \frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x \]
Substituting into the given expression:
\[ 5\cos x+3\left(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\right)+4 \]
\[ = 5\cos x+\frac{3}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+4 \]
\[ = \frac{13}{2}\cos x-\frac{3\sqrt{3}}{2}\sin x+4 \]
Now compare with
\[ a\cos x+b\sin x \]
where
\[ a=\frac{13}{2}, \qquad b=-\frac{3\sqrt{3}}{2} \]
Maximum value of \[ a\cos x+b\sin x \] is
\[ \sqrt{a^2+b^2} \]
So,
\[ \sqrt{ \left(\frac{13}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2 } \]
\[ = \sqrt{ \frac{169}{4} + \frac{27}{4} } \]
\[ = \sqrt{\frac{196}{4}} \]
\[ = \sqrt{49} =7 \]
Therefore,
Maximum value of the given expression:
\[ 7+4=11 \]
Minimum value of the given expression:
\[ -7+4=-3 \]
Final Answer
\[ \boxed{\text{Maximum value }=11} \]
\[ \boxed{\text{Minimum value }=-3} \]