Question

Find the distance between the following pairs of points -

(a cos A, a sin A) and (a cos B, a sin B)

Answer 1

pythagoras for this one (aka "distance formula")
\[d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

Answer 2

Right, but what do you get after substitution & simplification?

Answer 3

how about you post what you get after substitution ? (as a first step)

Answer 4

What I got is.....\[a.\sqrt{2(1-\cos(A-B))}\]
but the answer key shows.....
2a.sin((A-B)/2)

Answer 5

your answer is correct. But they used some trig identities to change it.

Do you know the half-angle formulas?
http://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php
often written as
\[ \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos(x)}{2}}\]
if we square both sides, we have
\[ \sin^2\left(\frac{x}{2}\right) = \frac{1-\cos(x)}{2}\\
2\sin^2\left(\frac{x}{2}\right) = 1-\cos(x)
\]
if we multiply both sides by 2 we get
\[ 4\sin^2\left(\frac{x}{2}\right) = 2(1-\cos(x)) \]
now take the square root of both sides
\[ 2 \sin\left(\frac{x}{2}\right) = \sqrt{2(1-\cos(x))}\]
use this result with x = A-B

Answer 6

Thank you Phi! shall brush up half angle formula...