If sin theta = frac{1}{2} left( sqrt{frac{x}{ | vedclass

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(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \ge

Vedclass · Accepted Answer

$x = y$

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(A) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \ge \sqrt{ab}$.Let $a = \sqrt{\frac{x}{y}}$ and $b = \sqrt{\frac{y}{x}}$.Then $\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight) \ge \sqrt{\sqrt{\frac{x}{y}} \cdot \sqrt{\frac{y}{x}}} = \sqrt{1} = 1$.Since $\sin 	heta = \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight)$ and we know that $\sin 	heta \le 1$ for all $	heta \in \mathbb{R}$,the only possible value for the expression is $1$.Therefore,$\frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} ight) = 1$.This implies $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = 2$.Squaring both sides,we get $\frac{x}{y} + \frac{y}{x} + 2 = 4$,which simplifies to $\frac{x}{y} + \frac{y}{x} = 2$.Multiplying by $xy$,we get $x^2 + y^2 = 2xy$,or $(x-y)^2 = 0$.Thus,$x = y$.

If $\sin \theta = \frac{1}{2} \left( \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} \right)$,where $x, y \in \mathbb{R} - \{0\}$. Then:

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