Question

What is the distance between the points
$P(m\cos (2\alpha ),m\sin (2\alpha ))$and
$Q(m\cos (2\beta ),\sin (2\beta ))$

1. $|2m\sin (\alpha - \beta )|$
2. $|2m\cos (\alpha - \beta )|$
3. $|m\sin (2\alpha - 2\beta )|$
4. $|m\sin (2\alpha + 2\beta )|$

Answer 1

Hint : This is a question involving multiple concepts from the topics like co-ordinate geometry and trigonometry. The concept of distance between points with certain trigonometric identities would help you solve this question.
$D = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$, where D is the distance between the points and
${\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1$, where $\theta$ can be any real number
$\cos (\alpha - \beta ) = \sin (\alpha )\sin (\beta ) + \cos (\alpha )\cos (\beta )$, where $\alpha$ and $\beta$ are real numbers
$\cos (2\theta ) = 1 - {\sin ^2}(\theta )$, where $\theta$ can be any real number

** Complete step-by-step answer** :
Now, let us start the question by applying the distance formula between the points $P(m\cos (2\alpha ),m\sin (2\alpha ))$and $Q(m\cos (2\beta ),\sin (2\beta ))$
$\Rightarrow D = \sqrt {{{(m\cos (2\alpha ) - m\cos (2\beta ))}^2} + {{(m\sin (2\alpha ) - m\sin (2\beta ))}^2}}$
Now, taking out m common from the bracket to obtain the equation shown below,
$\Rightarrow D = \sqrt {{m^2}{{(\cos (2\alpha ) - \cos (2\beta ))}^2} + {m^2}{{(\sin (2\alpha ) - \sin (2\beta ))}^2}}$
Now, taking m2 out of the square root function we get the equation shown below,
$\Rightarrow D = |m|\sqrt {{{(\cos (2\alpha ) - \cos (2\beta ))}^2} + {{(\sin (2\alpha ) - \sin (2\beta ))}^2}}$
Now, opening the brackets of the individual square terms as in the equation as shown below,
$\Rightarrow D = |m|\sqrt {{{\cos }^2}(2\alpha ) + {{\cos }^2}(2\beta ) - 2\cos (2\alpha )\cos (2\beta ) + {{\sin }^2}(2\alpha ) + {{\sin }^2}(2\beta ) - 2\sin (2\alpha )\sin (2\beta )}$
Now, grouping certain trigonometric terms to form identities as shown below,
$\Rightarrow D = |m|\sqrt {{{\cos }^2}(2\alpha ) + {{\sin }^2}(2\alpha ) + {{\cos }^2}(2\beta ) + {{\sin }^2}(2\beta ) - 2\cos (2\alpha )\cos (2\beta ) - 2\sin (2\alpha )\sin (2\beta )}$
Now, applying the identity ${\sin ^2}(\theta ) + {\cos ^2}(\theta ) = 1$as shown below,
$\Rightarrow D = |m|\sqrt {1 + 1 - 2\cos (2\alpha )\cos (2\beta ) - 2\sin (2\alpha )\sin (2\beta )}$
Now, taking -2 common from the terms as shown below,
$\Rightarrow D = |m|\sqrt {2 - 2(\cos (2\alpha )\cos (2\beta ) + \sin (2\alpha )\sin (2\beta ))}$
Now, let us again take 2 common in the equation,
$\Rightarrow D = |m|\sqrt {2 - 2(\cos (2\alpha - 2\beta ))}$
Now, let us apply the trigonometric identity $\cos (2\theta ) = 1 - {\sin ^2}(\theta )$ in the equation below,
$\Rightarrow D = |m|\sqrt {2(1 - \cos (2\alpha - 2\beta ))}$
$\Rightarrow D = |m|\sqrt {2(2{{\sin }^2}(\alpha - \beta ))}$
Now, after grouping everything we get,
$\Rightarrow D = |m|\sqrt {{{(4\sin (\alpha - \beta ))}^2}}$
Taking the square root on the right-hand side of the equation as shown below,
$\Rightarrow D = |m||2\sin (\alpha - \beta )|$
Now, we get the final answer as shown below,
$\Rightarrow D = |2m\sin (\alpha - \beta )|$
Therefore, option(1) is the correct answer.
So, the correct answer is “Option 1”.

Note : This question is based on the basic concepts of geometry and trigonometry. One should be well versed with them before solving this question. Do not commit calculation mistakes and be sure of the final answer.