Question

Find the distance between the points (cos 46,sin46) and (-sin 46, cos 46)

Answer 1

i think it seems to be better to follow this way

Answer 2

The distance between given two points (cos 46,sin 46) and (-sin 46, cos 46) is 1.414.

Step-by-step explanation:

Given:

Points are (cos 46,sin 46) and (-sin 46, cos 46).

To Find:

The distance between given two points (cos 46, sin 46) and (-sin 46, cos 46) is

Formula Used:

Point A coordinate is  (x 1 , y 1) and Point B coordinate is  (x 2,y 2). The distance between  points AB : $\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

Solution:

As given- Two points are (cos 46, sin 46) and (-sin 46, cos 46).

Let point A coordinate (x 1,y 1) =(cos 46,sin 46)

                              x1= cos46     ,  x2= sin46

and point B coordinate (x 2,y 2) =(-sin 46, cos 46)              

                             y1=-sin 46      ,  y2= cos 46

                             

Putting  values of x 1,x 2,y 1 and y 2 are in formula no.01.

The distance between given two points AB = $=\sqrt{(-sin48-cos48)^{2}+(cos48- sin48)^{2} }$

$=\sqrt{(sin48+cos48)^{2}+(cos48- sin48)^{2} }$

$=\sqrt{(sin^{2}48 +cos^{2} 48 +2 sin48 cos48)+(cos^{2}48+ sin^{2} 48 - 2cos48 sin48) }$

$=\sqrt{2(sin^{2}48 +cos^{2} 48 ) }$

Using identity $\(sin^{2}48 +cos^{2} 48 } =1$

$=\sqrt{2(1) }$

$=\sqrt{2 }$   =1.414

Thus, The distance between  two given points (cos 46,sin 46) and (-sin 46, cos 46) is 1.414