Trigonometry & Inverse Trigonometry Question - None

Question

Let $0 < \theta < \frac{\pi}{2}$ and $x = X \cos \theta + Y \sin \theta$, $y = X \sin \theta - Y \cos \theta$ such that $x^2 + 2xy + y^2 = aX^2 + bY^2$, where $a$ and $b$ are constants. Then

Accepted Answer

Step 1: We have $x^2 + y^2 = X^2 + Y^2$ (from rotation of coordinates)

Step 2: Calculate $xy = (X^2 - Y^2)\sin \theta \cos \theta - XY(\cos^2 \theta - \sin^2 \theta)$

Step 3: $x^2 + 2xy + y^2 = X^2 + Y^2 + 2(X^2 - Y^2)\sin 2\theta - 2XY \cos 2\theta$

Step 4: Expanding with $\sin 2\theta$ and $\cos 2\theta$ terms:

$x^2 + 2xy + y^2 = (1 + 2\sin 2\theta)X^2 + (1 - 2\sin 2\theta)Y^2 - 2\cos 2\theta \cdot XY$

Step 5: For this to equal $aX^2 + bY^2$ (with no cross term), we need $\cos 2\theta = 0$, giving $\theta = \frac{\pi}{4}$, hence $\sin 2\theta = 1$

Step 6: Therefore $a = 1 + 2 = 3$ and $b = 1 - 2 = -1$

∴ Answer is (c) $a = 3, b = -1$

Question:

Step-by-Step Solution

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