Question

For all $\theta$ in $\left[ {0,\dfrac{\pi }{2}} \right]$, then $\cos \left( {\sin \theta } \right)$ is
1)Less than $\sin \left( {\cos \theta } \right)$
2)Greater than $\sin \left( {\cos \theta } \right)$
3)Equal to $\sin \left( {\cos \theta } \right)$
4)None of these.

Answer 1

Here we will use the proper range and domain of the sine and cosine function. So, we will use these properties of sine and cosine function to simplify the problem and also we will use the given co domain in solving this problem and finally we will use the trigonometric functions to get the final result.

Complete step-by-step answer:
We will consider the sum of sine and cosine function i.e $\cos \theta + \sin \theta$.
Now, we will multiply and divide both the terms by the number $\sqrt 2$. Therefore, we get
$\Rightarrow \cos \theta + \sin \theta = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} \cdot \cos \theta + \dfrac{1}{{\sqrt 2 }} \cdot \sin \theta } \right)$
We know the values of the trigonometric functions: $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
Substituting $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ in the above equation, we get
$\Rightarrow \cos \theta + \sin \theta = \sqrt 2 \left( {\sin \dfrac{\pi }{4} \cdot \cos \theta + \cos \dfrac{\pi }{4} \cdot \sin \theta } \right)$
Now, we know from sum trigonometric identities that $\sin \left( {x + y} \right) = \sin x \cdot \cos y + \sin y \cdot \cos x$ .
Now, using the identity , in the above question, we get
$\Rightarrow \cos \theta + \sin \theta = \sqrt 2 \sin \left( {\dfrac{\pi }{4} + \theta } \right)$
Here, from the given condition, we know:
$\Rightarrow \sin \left( {\dfrac{\pi }{4} + \theta } \right) \le 1$
Now, we will multiply both sides of inequality by $\sqrt 2$. Therefore, we get
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{\pi }{4} + \theta } \right) \le \sqrt 2$ ………. $\left( 1 \right)$
We know that $\sqrt 2$ is less than $\dfrac{\pi }{2}$ i.e.
$\sqrt 2 < \dfrac{\pi }{2}$
So we can write equation $\left( 1 \right)$ as
$\Rightarrow \sqrt 2 \sin \left( {\dfrac{\pi }{4} + \theta } \right) < \dfrac{\pi }{2}$
Thus, we get the result as
$\Rightarrow \cos \theta + \sin \theta < \dfrac{\pi }{2}$
Subtracting $\sin \theta$ from both sides, we get
$\Rightarrow \cos \theta < \dfrac{\pi }{2} - \sin \theta$
Now, we will take sine function on both sides of the inequality. Thus, we get
$\Rightarrow \sin \left( {\cos \theta } \right) < \sin \left( {\dfrac{\pi }{2} - \sin \theta } \right)$
We know from periodic identity of trigonometry that $\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta$
Using this identity here, we get
$\Rightarrow \sin \left( {\cos \theta } \right) < \cos \left( {\sin \theta } \right)$
Thus, $\cos \left( {\sin \theta } \right)$ is greater than $\sin \left( {\cos \theta } \right)$.
Hence, the correct option is option 2.

Note: Here we have used trigonometric identities to simplify our equation. Trigonometric identities are defined as the equalities which involve the trigonometric functions. They are true for every value of the occurring variables for which both sides of the equality are defined. Also, all the trigonometric identities are periodic in nature as they repeat their values after a certain interval.