Question

The distance between the points $\left( {a\cos \alpha ,a\sin \alpha } \right)$ and $\left( {a\cos \beta ,a\sin \beta } \right)$ is
a. $a\cos \dfrac{{\alpha - \beta }}{2}$
b. $2a\cos \dfrac{{\alpha - \beta }}{2}$
c. $2a\sin \dfrac{{\alpha - \beta }}{2}$
d. $a\sin \dfrac{{\alpha - \beta }}{2}$

Answer 1

Hint: We will first calculate the distance between two points using the distance formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $, where $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are the coordinates of the points and $d$ 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, $\left( {a\cos \alpha ,a\sin \alpha } \right)$ and $\left( {a\cos \beta ,a\sin \beta } \right)$
As, it is known that if the coordinates of the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, then the distance between two points is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
On substituting the values, we will get,
$d = \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 ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$d = \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 ${a^2}$ and use the identity ${\sin ^2}x + {\cos ^2}x = 1$ to further simplify it
$
  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 $\cos a\cos b + \sin a\sin b = \cos \left( {a - b} \right)$ to simplify the expression.
$
  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, $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$
Therefore, we have,
$
  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,
\[ \Rightarrow d = 2a\sin \left( {\dfrac{{\alpha - \beta }}{2}} \right)\]
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.