Question

Let the angle $\theta \in \left[ {0,\dfrac{\pi }{2}} \right]$.which of the following is true?
\[
  A.{\text{ }}{\sin ^2}\theta > {\cos ^2}\theta \\
  B.{\text{ }}{\sin ^2}\theta < {\cos ^2}\theta \\
  C.{\text{ }}\sin \theta > \cos \theta \\
  D.{\text{ }}\cos \theta > \sin \theta \\
  E.{\text{ }}\sin \theta + \cos \theta \leqslant \sqrt 2 \\
 \]

Answer 1

Hint: In these types of questions where we need to find the relation between two trigonometric values for a given set of angles. There may be different true relations existing between the trigonometric terms. So we need to check for the true relations within the given set of options.

Complete step-by-step answer:

Given that $\theta \in \left[ {0,\dfrac{\pi }{2}} \right]$
As we know the relation between $\sin \theta {\text{ and }}\cos \theta $ for the angle between the given angle $\theta \in \left[ {0,\dfrac{\pi }{2}} \right]$ .
We know that for all the angles $\theta \in \left[ {0,\dfrac{\pi }{4}} \right]$ we have the relation$\cos \theta > \sin \theta $.
Also we know that for all the angles $\theta \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]$ we have the relation$\cos \theta < \sin \theta $ .
So the options C and D are incorrect.
As we have for all the angles $\theta \in \left[ {0,\dfrac{\pi }{4}} \right]$ we have the relation$\cos \theta > \sin \theta $. So, for all the angles $\theta \in \left[ {0,\dfrac{\pi }{4}} \right]$ we also have the relation ${\cos ^2}\theta > {\sin ^2}\theta $ .
And as have for all the angles $\theta \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]$ we have the relation$\cos \theta < \sin \theta $. So, for all the angles $\theta \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right]$ we also have the relation ${\cos ^2}\theta < {\sin ^2}\theta $ .
So, option A and B are also incorrect.
Now , we are left with just option E.
We have an unique relation between the sum of $\sin \theta {\text{ and }}\cos \theta $ that is \[ - \sqrt {{a^2} + {b^2}} \leqslant a\sin \theta + b\cos \theta \leqslant \sqrt {{a^2} + {b^2}} \]
Now in the above equation taking the coefficient of $\sin \theta {\text{ and }}\cos \theta $ that is $a\& b$ as 1.
We get the following relation.
\[
   \Rightarrow - \sqrt {{a^2} + {b^2}} \leqslant a\sin \theta + b\cos \theta \leqslant \sqrt {{a^2} + {b^2}} \\
   \Rightarrow - \sqrt {{1^2} + {1^2}} \leqslant \left( 1 \right)\sin \theta + \left( 1 \right)\cos \theta \leqslant \sqrt {{1^2} + {1^2}} \\
   \Rightarrow \sin \theta + \cos \theta \leqslant \sqrt {1 + 1} \\
   \Rightarrow \sin \theta + \cos \theta \leqslant \sqrt 2 \\
 \]
Hence, the relation E is correct.
So, E is the correct option.

Note: Whenever we come across the question with option, once try to solve the problem directly by putting in the values from the option. Students must remember the inequality relation between some important trigonometric terms at least for the first quadrant of the angular plane.