Question

What is the distance between `(4 ,( - 3pi)/8 )` and `(-1 ,( 3 pi )/4 )`?

Answer 1

`D=sqrt(17-8cos(pi/8))`

Explanation:

We know that ,

`"Distance between Polar Co-ordinates:"A(r_1,theta_1)and B(r_2,theta_2)` is

`color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)`

We have , `P_1(4,(-3pi)/8) and P_2(-1,(3pi)/4)`.

So , `r_1=4 , r_2=-1 , theta_1=(-3pi)/8 and theta_2=(3pi)/4`

`=>theta_1-theta_2=(-3pi)/8-(3pi)/4=(-3pi-6pi)/8=(-9pi)/8`

`=>cos(theta_1-theta_2)=cos((-9pi)/8)`
`=>cos(theta_1-theta_2)=cos((9pi)/8)to[becausecos(-theta)=costheta]`
`=>cos(theta_1-theta_2)=cos(pi+pi/8)=-cos(pi/8)`

`"Using : " color(red)((I)` we get

`D=sqrt(4^2+(-1)^2-2(4)(-1)(-cos(pi/8))`

`=>D=sqrt(16+1-8*cos(pi/8))`

`=>D=sqrt(17-8cos(pi/8))`

`=>D~~3`