Question

The biggest among $\left(\sin 1+\cos 1\right),\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right),\left(\sin 1-\cos 1\right):$
A) $\left(\sin 1+\cos 1\right)$
B) $\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right)$
C) $\left(\sin 1-\cos 1\right)$
D) None of these.

Answer 1

Hint: In trigonometric function, $\tan \theta$,
As angle approaches to zero ( $\theta \to 0$) , tangent of angle approached to infinity ($\tan \theta \to \mathrm{\infty }$).
$\Rightarrow \tan 1$must be a larger value (atleast greater than 1); use this fact to find inequality between $\sin 1\mathrm{\&}\cos 1$.
The Square of numbers for greater than 0 and less than 1, is smaller than the number.
For example: Square of 0.6
i.e. $${\left(0.6\right)}^2=0.36$$
0.36 < 0.6
The range is a set of all output values of function as independent variables varies thoughout the domain.
The domain is a set of all possible values on which function is defined.
For trigonometric function $$y=\sin x$$ , independent variable is x.
The domain is the set of all real numbers.
Range of trigonometric function $\sin x=\left[-1,1\right]\text{ and }\cos x=\left[-1,1\right]$
$$\begin{gathered} \Rightarrow {\left(\sin x\right)}^2<\sin x \\ \text{and }{\left(\cos x\right)}^2<\cos x \end{gathered}$$
Use above mention property to find the inequality between $\sqrt{\sin 1}\mathrm{\&}\sin 1$.

Complete step by step solution:
Step 1: Drawing a graph of the tangent function.

We know that $\tan 1>1$
It is known, $\tan 1={\displaystyle \frac{\sin 1}{\cos 1}}$
$\Rightarrow {\displaystyle \frac{\sin 1}{\cos 1}}>1$
$\Rightarrow \sin 1>\cos 1$ …… (1)
when smaller number is substracting form larger number, then result is real positive number.
$\Rightarrow \sin 1-\cos 1>0$ (from (1))
Hence, $\sin 1$is positive (or greater than 0) and $\cos 1$is also positive (or greater than 0).
$\Rightarrow \sin 1>0;\text{ cos1 > 0}$
The Sum of two positive numbers is greater than their difference.
$\Rightarrow \left(\sin 1+\cos 1\right)>\left(\sin 1-\cos 1\right)$ …… (2)
Step 2: Draw graph of the sine function

Range of sine function: $\sin x=\left[-1,1\right]$
$\Rightarrow \sin 1<1$
$\sqrt{\sin 1}>\sin 1$
Hence, for values less than ‘1’, higher power gives lower values
Example: For $\left(0<x<1\right)$
$x>x^2>x^3>x^4>x^5.....$
Similarly, $\sin 1>{\sin }^{2}1$
$$\begin{gathered} \Rightarrow \sin 1<\sqrt{\sin 1} \\ \mathrm{\&}\cos 1<\sqrt{\cos 1} \end{gathered}$$
Thus, $\left(\sqrt{\sin 1}-\sin 1\right)$ and $\left(\sqrt{\cos 1}-\cos 1\right)$ are real positive number.
Then, $\left(\sqrt{\sin 1}-\sin 1\right)+\left(\sqrt{\cos 1}-\cos 1\right)>0$
$\Rightarrow \left(\sqrt{\sin 1}+\sqrt{\cos 1}-\sin 1-\cos 1\right)>0$
On transferring number to the other side of the inequality, a sign of the number changes.
$\Rightarrow -\left(+\sin 1+\cos 1\right)>-\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right)$
Multiplying by a minus sign on both sides. The inequality reverses.
$\Rightarrow \left(\sin 1+\cos 1\right)<\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right)$ …… (3)
From (2) and (3)
$\left(\sin 1-\cos 1\right)<\left(\sin 1+\cos 1\right)<\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right)$
Final answer: Among $\left(\sin 1+\cos 1\right),\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right),\left(\sin 1-\cos 1\right)$; $\left(\sqrt{\sin 1}+\sqrt{\cos 1}\right)$is the biggest.

$\therefore$ The correct option is (B).

Note:
The range of a function is defined as the set of all values of the function defined on its domain.
Range and domain of some trigonometric functions are given below:

Trigonometric function	Domain	Range
Sine	$\left(-\mathrm{\infty },+\mathrm{\infty }\right)$	[-1,1]
Cosine	$\left(-\mathrm{\infty },+\mathrm{\infty }\right)$	[-1,1]
Tangent	All real numbers except ${\displaystyle \frac{\pi }{2}}+n\pi$	$\left(-\mathrm{\infty },+\mathrm{\infty }\right)$
Cosecant	All real numbers except $n\pi$	$$\left(-\mathrm{\infty },-1\right]\cup \left[1,+\mathrm{\infty }\right)$$
Secant	All real numbers except ${\displaystyle \frac{\pi }{2}}+n\pi$	$$\left(-\mathrm{\infty },-1\right]\cup \left[1,+\mathrm{\infty }\right)$$
Cotangent	All real numbers except $n\pi$	$\left(-\mathrm{\infty },+\mathrm{\infty }\right)$