Question

find the distance between the points p(cosΦ, sinΦ) ;q(sinΦ,-cosΦ)​

Answer 1

Answer:

equation of line joining the points (cos∅, sin∅) and (cosΦ, sinΦ) is

( y - sin∅) = (sinΦ - sin∅)/(cos∅ - cosΦ) (x -cos∅)

[ use formula , sinC - sinD = 2cos(C+D)/2.sin(C-D)/2 and CosC - cosD = -2sin(C + D)/2.sin(C-D)/2 ]

y - sin∅ = {2cos(∅+Φ)/2.sin(∅-Φ)/2/-2

[ use formula , sinC - sinD = 2cos(C+D)/2.sin(C-D)/2 and CosC - cosD = -2sin(C + D)/2.sin(C-D)/2 ]

y - sin∅ = {2cos(∅+Φ)/2.sin(∅-Φ)/2/-2sin(∅+Φ)/2.sin(∅-Φ)/2} (x - cos∅)

y - sin∅ =- {cos(∅+Φ)/2/sin(∅+Φ)/2}(x - cos∅)

{sin(∅+Φ)/2}y - sin∅.sin(∅+Φ)/2 + {cos(∅+Φ)/2}x - cos∅.cos(∅+Φ)/2 = 0

[ cos(A- B ) = cosA.cosB + sinA.sinB use it here ]

{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -cos(2∅-∅-Φ)/2 = 0

{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -{cos(∅-Φ)/2} =0

it distance from origin (0,0) is

P = | 0 + 0 - cos(∅ - Φ)/2|/√{cos²(∅+Φ)/2 + sin²(∅+Φ)/2 }

[ use formula, P = |ax1 + by1 +c|/√(a² + b²) ]

P = cos(∅ - Φ)/2

I hope you understand