Question

The distance between A(cosθ, sinθ) and B(−sinθ,cosθ) is​

Answer 1

Answer:

$ab \: = \sqrt{( - \sin\alpha - \cos\alpha) {}^{2} + ( \cos\alpha - \sin \alpha ) {}^{2} } \\ ab \: = \sqrt{ \sin {}^{2} \alpha + \cos {}^{2} \alpha + 2 \sin\alpha \cos\alpha \: + \cos {}^{2} \alpha + \sin {}^{2} \alpha - 2 \sin\alpha \cos\alpha } \\ since \: \sin {}^{2} \alpha + \cos {}^{2} \alpha = 1 \\ ab \: = \sqrt{1 + 2 \sin \alpha \cos\alpha + 1 - 2 \sin \alpha \cos\alpha } \\ ab \: = \sqrt{2} units \\ thus \: distance \: between \: a \: and \: b \: \: is \: \sqrt{2 \: } \: \: units \: \: \: \: ans$