First slide

Trigonometric equations

Question

$\sin θ = \frac{1}{2} (\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}})$ necessarily implies

Easy

A

$x < y$
B

$x > y$
C

$x = y$
D

$x$ and $y$

Solution

$S i n c e - 1 \leq \sin x \leq 1,$
$\Rightarrow - 1 \leq \frac{1}{2} (\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}) \leq 1$
$\Rightarrow - 2 \sqrt{x y} \leq x + y \leq 2 \sqrt{x y}$
$\Rightarrow x + y \leq 2 \sqrt{x y}$
$\Rightarrow x + y - 2 \sqrt{x y} \leq 0 \Rightarrow {(\sqrt{x} - \sqrt{y})}^{2} \leq 0 \Rightarrow {(\sqrt{x} - \sqrt{y})}^{2} = 0 (∵ s q u a r e i n n o n - n e g a t i v e)$
$\Rightarrow x = y$

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