The equation sin^2 theta = frac{x^2 + y^2}{2x | vedclass

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(A) Given the equation $\sin^2 	heta = \frac{x^2 + y^2}{2xy}$.Since $\sin^2 	heta \leq 1$,we must have $\frac{x^2 + y^2}{2xy} \leq 1$.Also,for $\sin

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$x = y$

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(A) Given the equation $\sin^2 	heta = \frac{x^2 + y^2}{2xy}$.Since $\sin^2 	heta \leq 1$,we must have $\frac{x^2 + y^2}{2xy} \leq 1$.Also,for $\sin^2 	heta$ to be non-negative,$x$ and $y$ must have the same sign,so $\frac{x^2 + y^2}{2xy} > 0$.By the Arithmetic Mean-Geometric Mean inequality,for positive $x, y$,$\frac{x^2 + y^2}{2} \geq \sqrt{x^2 y^2} = |xy|$.Thus,$\frac{x^2 + y^2}{2xy} \geq 1$ when $x, y$ have the same sign.Since we require $\frac{x^2 + y^2}{2xy} \leq 1$ and $\frac{x^2 + y^2}{2xy} \geq 1$,it must be that $\frac{x^2 + y^2}{2xy} = 1$.This implies $x^2 + y^2 = 2xy$,which simplifies to $(x - y)^2 = 0$,hence $x = y$.

The equation $\sin^2 \theta = \frac{x^2 + y^2}{2xy}$,where $x, y \neq 0$,is possible if

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