First slide

Extreme values and Periodicity of Trigonometric functions

Question

The difference of the max and the min values of the function $5 \cos x + 3 \cos (x + \frac{π}{3}) + 8$ over R is

Moderate

A

14.00
B

1
C

2
D

3

Solution

Let $f (x) = 5 \cos x + 3 \cos (x + 60 °) + 8$
$= 5 \cos x + 3 (\cos x \cos 60 ° - \sin x \sin 60 °) + 8$
$= \frac{13}{2} \cos x - \frac{3 \sqrt{3}}{2} \sin x + 8$
This is of the form $a \cos x + b \sin x + c,$
where $a = \frac{13}{2}, b = \frac{- 3 \sqrt{3}}{2}, c = 8$
We have $\sqrt{a^{2} + b^{2}} = 7$
The extreme values of f over R are
$c \pm \sqrt{a^{2} + b^{2}} = 8 \pm 7 = 15, 1$
‘15’ is the max value and ‘1' is the min value of the function f over R

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