First slide

Trigonometric Identities

Question

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

Easy

A

x=y
B

x=-y
C

2x=y
D

none of these

Solution

$Now, \sin^{2} θ = \frac{x^{2} + y^{2}}{2 xy}$
x, y have same sign
$Now \frac{x^{2} + y^{2}}{2 xy} = \frac{1}{2} [{(\sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}})}^{2} + 2] \geq 1$
But $\sin^{2} θ \leq 1 . Therefore, \frac{x^{2} + y^{2}}{2 xy} = 1$
x=y

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