First slide

Extreme values and Periodicity of Trigonometric functions

Question

The range of $\cos θ (\sin θ + \sqrt{\sin^{2} θ + \sin^{2} α})$ is

Difficult

A

$[- \sqrt{1 + \sin^{2} α}, \sqrt{1 + \sin^{2} α}]$
B

$[- \sqrt{1 + \cos^{2} α}, \sqrt{1 + \cos^{2} α}]$
C

$[\sin α, \cos α]$
D

no solution

Solution

$L e t f (θ) = \cos θ (\sin θ + \sqrt{\sin^{2} θ + \sin^{2} α})$
$\Rightarrow {(f (θ) \sec θ - \sin θ)}^{2} = \sin^{2} θ + \sin^{2} α$
$\Rightarrow f^{2} (θ) (1 + \tan^{2} θ) - 2 f (θ) \tan θ - \sin^{2} α = 0$
$\Rightarrow f^{2} (θ) \tan^{2} θ - 2 f (θ) \tan θ + (f^{2} (θ) - \sin^{2} α) = 0$
$∵ \tan θ i s r e a l, w e h a v e b^{2} - 4 a c \geq 0$
$\Rightarrow 4 f^{2} (θ) - 4 f^{2} (θ) (f^{2} (θ) - \sin^{2} α) \geq 0$
$\begin{array}{l} \Rightarrow 4 f^{2} (θ) (1 - f^{2} (θ) + \sin^{2} α) \geq 0 \\ \Rightarrow 1 - f^{2} (θ) + \sin^{2} α \geq 0 \\ \Rightarrow f^{2} (θ) \leq 1 + \sin^{2} α \\ \Rightarrow - \sqrt{1 + \sin^{2} α} \leq f (θ) \leq \sqrt{1 + \sin^{2} α} \end{array}$

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