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

प्रश्न

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

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

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उत्तर

$Let f (x) = 5 \cos x + 3 \sin (\frac{π}{6} - x) + 4$
$Now f (x) = 5 \cos x + 3 (\sin 30 ° \cos x - \cos 30 ° \sin x) + 4$
$= 5 \cos x + \frac{3}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x + 4$
$= \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x + 4$
$We know that$
$- \sqrt{{(\frac{13}{2})}^{2} + {(- \frac{3 \sqrt{3}}{2})}^{2}} \leq \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{{(\frac{13}{2})}^{2} + {(- \frac{3 \sqrt{3}}{2})}^{2}} for all x$
$Therefore,$
$- \sqrt{\frac{169 + 27}{4}} \leq \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{\frac{169 + 27}{4}}$
$\Rightarrow - \frac{14}{2} + 4 \leq \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x + 4 \leq \frac{14}{2} + 4$
$\Rightarrow - 3 \leq \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x + 4 \leq 11$
$Hence, maximum and minimun values of f (x) are 11 and - 3, respectively .$

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Trigonometric Functions of Sum and Difference of Two Angles

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.2 [पृष्ठ २६]

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अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.2 | Q 1.3 | पृष्ठ २६

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

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

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