Question

the distance between points (cosQ,sinQ ) (cos p, sin p)​

Answer 1

Answer:

$\sqrt{2-2cos(Q-p)}$

Step-by-step explanation:

 (cosQ , sinQ)   and(cosp , sinp)

if the points are (x1,y1)  and (x2,y2) then

Distance between any points =$\sqrt{(x1-x2)^2+(y1-y2)^2}$

step 1

substitute the points in the theorem

     Distance = $\sqrt{(cosQ-cosp)^2+(sinQ-sinp)^2}$

step 2

    expand each of complete square

       Distance = $\sqrt{(cos^{2}Q-2cosQcosp+cos^{2}p)+( Sin^2Q-2sinQsinp+sin^2p) }$

step 3

       use    $cos^{2} x+sin^2x=1$  and simplify

$\sqrt{cos^2Q+sin^2Q+cos^2p+sin^2p - 2cosQcop-2sinQsinp}$

$cos^2Q+sin^2Q =1 and cos^2p+sin^2p =1$

$\sqrt{1+1-2cosQcosp-2sinQsinp}$

$\sqrt{2-2cosQcosp-2sinQsinp}$

step 4

using trigonometry and make a simplification of the expression like

$cosQcosp+sinQsinp =Cos(Q-p)$

$\sqrt{2-2(cosQcosp+sinQsinp)}$

$\sqrt{2-2cos(Q-p)}$