Question

The distance between the points (a cos \(\theta\) + b sin \(\theta\), 0) and (0, a sin \(\theta\) – b cos \(\theta\)) is

• A. \(\sqrt{a^2 + b^2}\)
• B. \(a^2 - b^2\)
• C. \(\sqrt{a^2 - b^2}\)
• D. \(a^2 + b^2\)

Hint:

For any perpendicular coordinate components of the form \((x, 0)\) and \((0, y)\), the distance is always \(\sqrt{x^2 + y^2}\).
Since the sum of the squares of these specific trigonometric terms always simplifies to \(a^2 + b^2\), you can write down the answer directly as \(\sqrt{a^2 + b^2}\)!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic of this question is Coordinate Geometry.
We are given two points in the Cartesian plane with coordinates containing trigonometric functions.
We need to find the distance between these two points and simplify the resulting expression using trigonometric identities.

Step 2: Key Formula or Approach:
We will use the standard distance formula to calculate the distance between the points \((x_1, y_1)\) and \((x_2, y_2)\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] We will substitute the coordinates of the points, expand the squared terms using algebraic identities, and simplify using the fundamental Pythagorean identity:
\[ \sin^2\theta + \cos^2\theta = 1 \]

Step 3: Detailed Explanation:

• Identify the coordinates of the given points:
\[ (x_1, y_1) = (a\cos\theta + b\sin\theta, 0) \] \[ (x_2, y_2) = (0, a\sin\theta - b\cos\theta) \]

• Substitute these coordinates into the distance squared formula:
\[ d^2 = (0 - (a\cos\theta + b\sin\theta))^2 + ((a\sin\theta - b\cos\theta) - 0)^2 \] \[ d^2 = (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 \]

• Expand both terms using the algebraic identity \((x \pm y)^2 = x^2 \pm 2xy + y^2\):
- Expand the first term:
\[ (a\cos\theta + b\sin\theta)^2 = a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta \] - Expand the second term:
\[ (a\sin\theta - b\cos\theta)^2 = a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta \]

• Add the two expanded expressions together:
Notice that the cross-product terms cancel out:
\[ 2ab\sin\theta\cos\theta - 2ab\sin\theta\cos\theta = 0 \] Thus, the equation simplifies to:
\[ d^2 = a^2\cos^2\theta + b^2\sin^2\theta + a^2\sin^2\theta + b^2\cos^2\theta \]

• Group the terms containing \(a^2\) and \(b^2\) together:
\[ d^2 = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta) \]

• Substitute the trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\) into the equation:
\[ d^2 = a^2(1) + b^2(1) \] \[ d^2 = a^2 + b^2 \]

• Take the square root on both sides to find the distance \(d\):
\[ d = \sqrt{a^2 + b^2} \]

Step 4: Final Answer:
The distance between the two points is \(\sqrt{a^2 + b^2}\).
Therefore, the correct option is (A).