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Question: The distance between the points \(\left( {a\cos \alpha ,a\sin \alpha } \right)\) and \(\left( {a\cos...

The distance between the points (acos⁡α,asin⁡α)\left( {a\cos \alpha ,a\sin \alpha } \right) and (acos⁡β,asin⁡β)\left( {a\cos \beta ,a\sin \beta } \right) is
a. acos⁡α−β2a\cos \dfrac{{\alpha - \beta }}{2}
b. 2acos⁡α−β22a\cos \dfrac{{\alpha - \beta }}{2}
c. 2asin⁡α−β22a\sin \dfrac{{\alpha - \beta }}{2}
d. asin⁡α−β2a\sin \dfrac{{\alpha - \beta }}{2}

Explanation

Solution

We will first calculate the distance between two points using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} , where (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) are the coordinates of the points and dd is the distance between these two points. Further, simplify the expression using the trigonometric identities to get the required answer.

Complete step-by-step answer:
We are given the coordinates of two points, (acos⁡α,asin⁡α)\left( {a\cos \alpha ,a\sin \alpha } \right) and (acos⁡β,asin⁡β)\left( {a\cos \beta ,a\sin \beta } \right)
As, it is known that if the coordinates of the points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right), then the distance between two points is d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
On substituting the values, we will get,
d=(acos⁡β−acos⁡α)2+(asin⁡β−asin⁡α)2d = \sqrt {{{\left( {a\cos \beta - a\cos \alpha } \right)}^2} + {{\left( {a\sin \beta - a\sin \alpha } \right)}^2}}
Simplify the brackets using the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
d=a2cos⁡2α+a2cos⁡2β−2a2cos⁡αcos⁡β+a2sin⁡2β+a2sin⁡α−2a2sin⁡αsin⁡βd = \sqrt {{a^2}{{\cos }^2}\alpha + {a^2}{{\cos }^2}\beta - 2{a^2}\cos \alpha \cos \beta + {a^2}{{\sin }^2}\beta + {a^2}\sin \alpha - 2{a^2}\sin \alpha \sin \beta }
Take a2{a^2} and use the identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 to further simplify it
d=a2(cos⁡2α+cos⁡2β−2cos⁡αcos⁡β+sin⁡2β+sin⁡2α−2sin⁡αsin⁡β) ⇒d=a2(1+1−2cos⁡αcos⁡β−2sin⁡αsin⁡β) ⇒d=a2(2−2cos⁡αcos⁡β−2sin⁡αsin⁡β) ⇒d=2a2(1−cos⁡αcos⁡β−sin⁡αsin⁡β)  d = \sqrt {{a^2}\left( {{{\cos }^2}\alpha + {{\cos }^2}\beta - 2\cos \alpha \cos \beta + {{\sin }^2}\beta + {{\sin }^2}\alpha - 2\sin \alpha \sin \beta } \right)} \\\ \Rightarrow d = \sqrt {{a^2}\left( {1 + 1 - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta } \right)} \\\ \Rightarrow d = \sqrt {{a^2}\left( {2 - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta } \right)} \\\ \Rightarrow d = \sqrt {2{a^2}\left( {1 - \cos \alpha \cos \beta - \sin \alpha \sin \beta } \right)} \\\
Now, we will take minus sign common and use the formula cos⁡acos⁡b+sin⁡asin⁡b=cos⁡(a−b)\cos a\cos b + \sin a\sin b = \cos \left( {a - b} \right) to simplify the expression.
d=2a2(1−(cos⁡αcos⁡β+sin⁡αsin⁡β)) ⇒d=2a2(1−cos⁡(α−β))  d = \sqrt {2{a^2}\left( {1 - \left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)} \right)} \\\ \Rightarrow d = \sqrt {2{a^2}\left( {1 - \cos \left( {\alpha - \beta } \right)} \right)} \\\
Also, 1−cos⁡x=2sin⁡2x21 - \cos x = 2{\sin ^2}\dfrac{x}{2}
Therefore, we have,
d=2a2(1−cos⁡(α−β)) ⇒d=2a2(2sin⁡2(α−β2)) ⇒d=4a2sin⁡2(α−β2)  d = \sqrt {2{a^2}\left( {1 - \cos \left( {\alpha - \beta } \right)} \right)} \\\ \Rightarrow d = \sqrt {2{a^2}\left( {2{{\sin }^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)} \right)} \\\ \Rightarrow d = \sqrt {4{a^2}{{\sin }^2}\left( {\dfrac{{\alpha - \beta }}{2}} \right)} \\\
Now, solving the square root of the above equation, we will get,
⇒d=2asin⁡(α−β2)\Rightarrow d = 2a\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)
Hence, option C is correct.

Note: Students should learn trigonometric identities and formulas to do these types of questions. Also, do not forget to simplify the equation by taking out common terms and then using trigonometric identities. The concept used in this question is simple but students often make mistakes in simplification of the distance calculated.