Question

The distance between the points (sin x , cos x) and ( cos x, -sin x) is
A. 1
B. √2
C. 2sin(x) ∙ cos(x)
D. 4sin(x) ∙ cos(x)
E. None of these

Answer 1

Hint: Start by considering the given two points as ${\text{P}}({x_1},{y_1}){\text{ and Q}}\left( {{x_2},{y_2}} \right)$ . Compare the coordinates for values and apply the formula for distance between two given points i.e. $s = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Simplify the values by using trigonometric identities ,to get the desired value.

Complete step by step answer:
Given
A(sin x , cos x) and B (cos x , -sin x)
We know that distance between two points ${\text{P}}({x_1},{y_1}){\text{ and Q}}\left( {{x_2},{y_2}} \right)$ is given by the formula $s = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
We have $[{x_1} = {\text{ }}sin{\text{ }}x{\text{ , }}{y_1} = cos{\text{ }}x]
[{x_2} = {\text{ }}cos{\text{ }}x{\text{ , }}{y_2} = {\text{ }} - sin{\text{ }}x]$
Substituting these values in the formula we get
$s = \sqrt {{{(\cos x - \sin x)}^2} + {{(( - \sin x) - \cos x)}^2}} $
By using the formula
$
  {(a - b)^2} = {a^2} + {b^2} - 2ab{\text{ and }}{(a + b)^2} = {a^2} + {b^2} + 2ab \\
   = \sqrt {{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x + {{( - \sin x)}^2} + {{(\cos x)}^2} - 2( - \sin x)(\cos x)} \\
   = \sqrt {1 - 2\sin x\cos x + {{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \\
   = \sqrt {1 + 1} \\
  S = \sqrt 2 \\
$
So, the correct answer is “Option B”.

Note: All the conversion and identities related to trigonometry must be well known in order to solve such similar questions. Attention is to be given while substituting the values of coordinates with proper sign (positive or negative).