Question

\[\left( {\sin \theta ,\cos \theta } \right)\] and (3, 2) lies on the same side of the line x + y = 1, then \[\theta \] lies between
A. \[\left( {0,\dfrac{\pi }{2}} \right)\]
B. \[\left( {0,\pi } \right)\]
C. \[\left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)\]
D. \[\left( {0,\dfrac{\pi }{4}} \right)\]

Answer 1

Hint: Theta \[\theta \] is the angle in radians which tells the value of a point \[\left( {\sin \theta ,\cos \theta } \right)\] for a particular value of theta in the range [-1,1]. When points lie on the same side of the line then it can be either the left side or the right side irrespective of the quadrant.


Formula Used: \[\sin A\cos B + \sin B\cos A = \sin \left( {A + B} \right)\] and \[{L_1}{L_2} > 0\], where \[{L_1}\] and \[{L_2}\] are two different equations of a line.


Complete step-by-step answer: We have been given a line as x + y = 1. We will substitute the point (3, 2) in the line x + y - 1 and check the result, 3 + 2 – 1 = 4.
Here, our result is greater than zero, that is, 4 > 0.
We will substitute the point \[\left( {\sin \theta ,\cos \theta } \right)\] in the line x + y - 1 and check the result, \[\sin \theta + \cos \theta - 1\].
As our first result of the line was greater the zero, so even here our result should be greater than zero, that is, \[\sin \theta + \cos \theta - 1 > 0\].
For the point \[\left( {\sin \theta ,\cos \theta } \right)\] and (3, 2) on the same side of the line x + y = 1, the product of the results of both equations should also be greater than zero. Let, \[{L_1} = 4\] and \[{L_2} = \sin \theta + \cos \theta - 1\].
\[
  {L_1}{L_2} > 0 \\
   \Rightarrow 4\left( {\sin \theta + \cos \theta - 1} \right) > 0 \\
   \Rightarrow \sin \theta + \cos \theta > 1 \\
   \Rightarrow \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin \theta + \dfrac{1}{{\sqrt 2 }}\cos \theta } \right) > 1
 \]
Further Simplifying, we get,
\[
  \cos \left( {\dfrac{\pi }{4}} \right)\sin \theta + \sin \left( {\dfrac{\pi }{4}} \right)\cos \theta > \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow \sin \left( {\theta + \dfrac{\pi }{4}} \right) > \dfrac{1}{{\sqrt 2 }} \\
   \Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\theta + \dfrac{\pi }{4}} \right)} \right) > {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow \theta + \dfrac{\pi }{4} > \dfrac{\pi }{4}
 \]
Since \[\sin \left( {\theta + \dfrac{\pi }{4}} \right)\] should be positive, so we can say,
\[
  \sin \left( {\theta + \dfrac{\pi }{4}} \right) < 1 \\
   \Rightarrow {\sin ^{ - 1}}\left( {\sin \left( {\theta + \dfrac{\pi }{4}} \right)} \right) < {\sin ^{ - 1}}\left( 1 \right) \\
   \Rightarrow \theta + \dfrac{\pi }{4} < \dfrac{\pi }{2}
 \]
Combining the two inequalities, we get,
\[
  \dfrac{\pi }{4} < \theta + \dfrac{\pi }{4} < \dfrac{\pi }{2} \\
   \Rightarrow \dfrac{\pi }{4} - \dfrac{\pi }{4} < \theta + \dfrac{\pi }{4} - \dfrac{\pi }{4} < \dfrac{\pi }{2} - \dfrac{\pi }{4} \\
   \Rightarrow 0 < \theta < \dfrac{\pi }{4}
 \]


So, option D, \[\left( {0,\dfrac{\pi }{4}} \right)\] is the required solution.

Note: Since the points \[\left( {\sin \theta ,\cos \theta } \right)\] and (3, 2) lies on the same side of the line x + y = 1, and the point (3, 2) is positive, so we can say that the point \[\left( {\sin \theta ,\cos \theta } \right)\] will also be positive. \[\sin \theta \] and \[\cos \theta \] are only positive for the values of \[\theta \] in the range \[\left( {0,\dfrac{\pi }{2}} \right)\]. So, \[\sin \left( {\theta + \dfrac{\pi }{4}} \right)\] should also be positive for \[\theta \] to lie in the range \[\left( {0,\dfrac{\pi }{2}} \right)\].