Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Maximum and Minimum Values of Each of the Following Trigonometrical Expression: 5 Cos X + 3 Sin ( π 6 − X ) + 4 - Mathematics

Short Note

Find the maximum and minimum values of each of the following trigonometrical expression:

$5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4$

#### Solution

$\text{ Let } f\left( x \right) = 5 \cos x + 3 \sin\left( \frac{\pi}{6} - x \right) + 4$
$\text{ Now } f\left( x \right) = 5\cos x + 3\left( \sin30°\cos x - \cos30°\sin x \right) + 4$
$= 5\cos x + \frac{3}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4$
$= \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4$
$\text{ We know that }$
$- \sqrt{\left( \frac{13}{2} \right)^2 + \left( - \frac{3\sqrt{3}}{2} \right)^2} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\left( \frac{13}{2} \right)^2 + \left( - \frac{3\sqrt{3}}{2} \right)^2} \text{ for all x }$
$\text{ Therefore },$
$- \sqrt{\frac{169 + 27}{4}} \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x \leq \sqrt{\frac{169 + 27}{4}}$
$\Rightarrow - \frac{14}{2} + 4 \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4 \leq \frac{14}{2} + 4$
$\Rightarrow - 3 \leq \frac{13}{2}\cos x - \frac{3\sqrt{3}}{2}\sin x + 4 \leq 11$
$\text{ Hence, maximum and minimun values of } f\left( x \right) \text{ are 11 and - 3, respectively } .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.2 | Q 1.3 | Page 26