First slide

Extreme values of trigonometric functions in trigonometry

Question

Let $f (x) = absin x + b \sqrt{1 - a^{2}} \cos x + c$ where $| a | < 1$ ,b > 0, then

Difficult

A

maximum value of f(x) is b if c = 0
B

difference of maximum and minimum values of f(x) is 2b
C

$f (x) = c if x = - \cos^{- 1} a$
D

$f (x) = c if x = \cos^{- 1} a$

Solution

$\begin{array}{l} f (x) = absin x + b \sqrt{1 - a^{2}} \cos x + c, where | a | < 1, b < 0 \\ \begin{aligned} f (x) & = \sqrt{a^{2} b^{2} + b^{2} - b^{2} a^{2}} \sin (x + α) + c \\ = bsin (x + α) + c, where \tan α = \frac{b \sqrt{1 - a^{2}}}{ab} = \frac{\sqrt{1 - a^{2}}}{a} \\ = bcos (x - α) + c, where \tan α = \frac{ab}{b \sqrt{1 - a^{2}}} = \frac{a}{\sqrt{1 - a^{2}}} \end{aligned} \\ f (x)_{max} - f (x)_{min} = c + b - (c - b) = 2 b \\ f (x) = c if x + α = 0 \end{array}$
$\begin{array}{l} or x = - α \\ or x = - \cos^{- 1} a \end{array}$

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