Question

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

Answer 1

Here you go......
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Answer 2

The distance of the point (a$\cos \alpha$, a$\sin \alpha$) from the origin is independent of $\alpha$, proved.

Explanation:

Given,

The distance of the point (a$\cos \alpha$, a$\sin \alpha$) from the origin (0, 0).

To prove that, the distance of the point (a$\cos \alpha$, a$\sin \alpha$) from the origin is independent of $\alpha$.

We know that,

The distance formula( between two points)

= $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

= $\sqrt{(0-a\cos \alpha)^{2}+(0-a\sin \alpha)^{2}}$

= $\sqrt{(-a\cos \alpha)^{2}+(-a\sin \alpha)^{2}}$

= $\sqrt{a^2\cos^{2} \alpha+a^2\sin^{2} \alpha$

Taking $a^2$ as common, we get

= $\sqrt{a^2(\cos^{2} \alpha+\sin^{2} \alpha)$

Using the trigonometric identity,

$\sin^{2} A+\cos^{2} A$ = 1

= $\sqrt{a^2(1)$

= $\sqrt{a^2$

= a units, it is independent of $\alpha$, proved.

Thus, the distance of the point (a$\cos \alpha$, a$\sin \alpha$) from the origin is independent of $\alpha$, proved.