Question

Find the distance ( a cos alpha , a sin alpha ) and (a cos beta , asin beta)

Answer 1

$Let \: A(x_{1} , y_{1}) = ( a cos \alpha , a sin \alpha ) \: and \\ B(x_{2} , y_{2}) = ( a cos \beta , a sin \beta )$

$\underline { \blue { By \: Distance \: Formula : }}$

$\boxed { \pink { AB = \sqrt{ (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} }}$

$AB = \sqrt{ ( a cos \beta - a cos \alpha )^{2} + ( a sin \beta - a sin \alpha )^{2} }$

$\implies AB = \sqrt{ a^{2} ( cos \beta - cos \alpha )^{2} + a^{2} ( sin \beta - sin \alpha )^{2} }$

$\implies AB =a \sqrt{ ( cos \beta - cos \alpha )^{2} + ( sin \beta - sin \alpha )^{2} }$

$\implies AB =a \sqrt{ cos^{2} \beta + cos^{2} \alpha -2 cos\alpha cos \beta + sin^{2}\beta + sin^{2} \alpha - 2 sin \alpha sin \beta }$

$\implies AB =a \sqrt{ (cos^{2} \beta + sin^{2} \beta) -2( cos\alpha cos \beta + sin \alpha sin \beta)+ (cos^{2}\alpha + sin^{2} \alpha) }$

$\implies AB =a \sqrt{ 1 -2( cos\alpha cos \beta + sin \alpha sin \beta)+ 1 }$

$\implies AB =a \sqrt{ 2 -2cos(\alpha - \beta )}$

$\implies AB =a \sqrt{ 2[1 -cos(\alpha - \beta )]}$

Therefore.,

$\red { Distance }\green {= a \sqrt{ 2[1 -cos(\alpha - \beta )]}}$

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