Question

The distance between (a cos alpha , a sin beta) and (a cos beta , a sin alpha) is
Answer the problem with full steps

Answer 1

let the distance between  A (x1,y1) and B (x2,y2) 
= √(x2-x1)² +(y2-y1)²

here A (a cos α, a sin β) = (x1,y1)
B(a cosβ, a sin α) = (x2,y2)

AB = √(acos β - a cos α)² + ( a sin α - a sin β)²

=√a²cos² β + a² cos² α -2a² cosαcosβ +a²sin²α + a² sin² β - 2a² sin α sin β

= √a²(cos²β +sin² β) + a²(cos² α +sin² α) -2a²(cosαcosβ+sinαsinβ)
= √a²+a² -2a² cos(α-β)
= √2a² -2a² cos(α-β)
= √2a²[1-cos(α-β)]
=a √2[1-cos(α-β)]

Answer 2

given,

(a cosx, a siny) (a cosy, a sinx)

Distance = √(acosy - acosx)²+(asinx-asiny)²

= √(a²cos²y+a²cos²x-2(acosy)(acosx))+(a²sin²x+a²sin²y-2(asinx)(asiny))

= √a²cos²y+a²cos²x-2a²cosxcosy+a²sin²x+a²sin²y-2a²sinxsiny

= √a²cos²x+a²sin²x+a²cos²y+a²sin²y -2a²cosxcosy-2a²sinxsiny

= √a²(cos²x+sin²x)+a²(cos²y+sin²y)-2a²(cosxcosy+sinxsiny)

= √a²+a²-2a²(cos(x-y))

= √2a²-2a²(cos(x-y))

= √2a²(1-cos(x-y))

= √2a²(2sin²(x-y/2))

= √4a²sin²(x-y/2)

= 2asin(x-y/2)

Used identities:

1)(a-b)²=a²+b²-2ab

2) sin²x+cos²x=1

3)cos(A-B) = cosAcosB+sinAsinB

4) sin(a/2) = √(1-cosa)/√2

squaring on both sides

=>sin²(a/2)=1-cosa/2

=>2sin²(a/2)=1-cosa

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