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Question: The distance between the points \((a\cos\alpha,a\sin\alpha)\) and \((a\cos\beta,a\sin\beta)\) is....

The distance between the points (acos⁡α,asin⁡α)(a\cos\alpha,a\sin\alpha) and (acos⁡β,asin⁡β)(a\cos\beta,a\sin\beta) is.

A

acos⁡α−β2a\cos\frac{\alpha - \beta}{2}

B

2acos⁡α−β22a\cos\frac{\alpha - \beta}{2}

C

asin⁡α−β2a\sin\frac{\alpha - \beta}{2}

D

2asin⁡α−β22a\sin\frac{\alpha - \beta}{2}

Answer

2asin⁡α−β22a\sin\frac{\alpha - \beta}{2}

Explanation

Solution

Distance (nwjh) =a2(cos⁡α−cos⁡β)2+a2(sin⁡α−sin⁡β)2= \sqrt { a ^ { 2 } ( \cos \alpha - \cos \beta ) ^ { 2 } + a ^ { 2 } ( \sin \alpha - \sin \beta ) ^ { 2 } }

=asin⁡2α+cos⁡2α+cos⁡2β+sin⁡2β−2cos⁡αcos⁡β−2sin⁡αsin⁡β= a \sqrt { \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha + \cos ^ { 2 } \beta + \sin ^ { 2 } \beta - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta }

=a2{1−cos⁡(α−β)}=2asin⁡(α−β2)= a \sqrt { 2 \{ 1 - \cos ( \alpha - \beta ) \} } = 2 a \sin \left( \frac { \alpha - \beta } { 2 } \right)

Trick : Put a=1,α=π2,β=π6a = 1 , \alpha = \frac { \pi } { 2 } , \beta = \frac { \pi } { 6 } then the points will be (0, 1) and (32,12)\left( \frac { \sqrt { 3 } } { 2 } , \frac { 1 } { 2 } \right). Obviously, the distance between these two points is 1 which is given by (4).