Prove that the distance of the point `(a cos alpha, a sin alpha)` from the origin is independent of `alpha`

Answer

To prove that the distance of the point \((a \cos \alpha, a \sin \alpha)\) from the origin is independent of \(\alpha\), we can follow these steps:

### Step 1: Identify the Points
We have the point \(P(a \cos \alpha, a \sin \alpha)\) and the origin \(O(0, 0)\).

### Step 2: Use the Distance Formula
The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In our case, we will use:
- \(x_1 = 0\), \(y_1 = 0\) (coordinates of the origin)
- \(x_2 = a \cos \alpha\), \(y_2 = a \sin \alpha\) (coordinates of point P)

### Step 3: Substitute the Coordinates into the Distance Formula
Substituting the coordinates into the distance formula, we get:
\[ d = \sqrt{(a \cos \alpha - 0)^2 + (a \sin \alpha - 0)^2} \]
This simplifies to:
\[ d = \sqrt{(a \cos \alpha)^2 + (a \sin \alpha)^2} \]

### Step 4: Factor Out \(a^2\)
We can factor out \(a^2\) from the expression inside the square root:
\[ d = \sqrt{a^2 (\cos^2 \alpha + \sin^2 \alpha)} \]

### Step 5: Use the Pythagorean Identity
We know from trigonometry that:
\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]
Substituting this identity into our expression gives:
\[ d = \sqrt{a^2 \cdot 1} = \sqrt{a^2} \]

### Step 6: Simplify the Expression
This simplifies to:
\[ d = a \]

### Conclusion
Thus, we have shown that the distance \(d\) from the point \((a \cos \alpha, a \sin \alpha)\) to the origin is:
\[ d = a \]
This distance is independent of \(\alpha\), as it does not contain \(\alpha\) in its expression.