CASE-BASED QUESTIONS
DOMAIN AND RANGE OF TRIGONOMETRIC FUNCTIONS

\begin{tabular}{|l|l|l|}
\hline
FUNCTION & DOMAIN & RANGE \\
\hline
[tex]$\sin x$[/tex] & [tex]$R$[/tex] & [tex]$[-1,1]$[/tex] \\
\hline
[tex]$\cos x$[/tex] & [tex]$R$[/tex] & [tex]$[-1,1]$[/tex] \\
\hline
[tex]$\tan x$[/tex] & [tex]$R - \left(2n+1\right) \frac{\pi}{2}$[/tex] & [tex]$R$[/tex] \\
\hline
[tex]$\csc x$[/tex] & [tex]$R - n\pi$[/tex] & [tex]$R \setminus (-1, 1)$[/tex] \\
\hline
[tex]$\sec x$[/tex] & [tex]$R - \left(2n+1\right) \frac{\pi}{2}$[/tex] & [tex]$R \setminus (-1, 1)$[/tex] \\
\hline
[tex]$\cot x$[/tex] & [tex]$R - n\pi$[/tex] & [tex]$R$[/tex] \\
\hline
\end{tabular}

Based on the above information, answer the following questions:

(i) Domain of [tex]$f(x)=\frac{1}{\sqrt{1+\cos x}}$[/tex] is:
(a) [tex]$R$[/tex]
(b) [tex]$R - (2n+1)\pi$[/tex]
(c) [tex]$R - 2n\pi$[/tex]
(d) [tex]$R - 1$[/tex]

(ii) Domain of [tex]$f(x)=\frac{1}{\sin x + \cos x}$[/tex] is:
(a) [tex]$R - \left(4n-1\right) \frac{\pi}{4}$[/tex]
(b) [tex]$R - \left(4n+1\right) \frac{\pi}{4}$[/tex]
(c) [tex]$R - \frac{n\pi}{4}$[/tex]
(d) [tex]$R - n\pi$[/tex]

(iii) Range of [tex]$f(x)=\sqrt{5+4\cos x}$[/tex] is:



Answer :

Sure, let's go through each part of the question step by step.

### (i) Domain of [tex]\( f(x) = \frac{1}{\sqrt{1 + \cos(x)}} \)[/tex]

To determine the domain of this function, we need to ensure the denominator is not zero and the expression under the square root is non-negative:
- The denominator [tex]\( \sqrt{1 + \cos(x)} \)[/tex] must be non-zero. This means [tex]\( 1 + \cos(x) \neq 0 \)[/tex].
- [tex]\( 1 + \cos(x) \)[/tex] is non-zero when [tex]\( \cos(x) \neq -1 \)[/tex].

The cosine function [tex]\(\cos(x)\)[/tex] equals [tex]\(-1\)[/tex] at [tex]\( x = (2n+1)\pi \)[/tex] where [tex]\( n \)[/tex] is an integer.

Therefore, the domain of the function [tex]\( \frac{1}{\sqrt{1+\cos(x)}} \)[/tex] is [tex]\( R - (2n+1)\pi \)[/tex].

Answer: (b) [tex]\( R - (2n+1)\pi \)[/tex]

### (ii) Domain of [tex]\( f(x) = \frac{1}{\sin(x) + \cos(x)} \)[/tex]

For the function to be defined, the denominator [tex]\(\sin(x) + \cos(x)\)[/tex] must not be zero. We need to find when [tex]\(\sin(x) + \cos(x) = 0\)[/tex]:
- This equation holds true when [tex]\( \sin(x) = -\cos(x) \)[/tex].

The expression [tex]\(\sin(x) = -\cos(x)\)[/tex] is true at [tex]\( x = (4n+1)\frac{\pi}{4} \)[/tex] where [tex]\( n \)[/tex] is an integer.

Therefore, the domain of the function [tex]\( \frac{1}{\sin(x) + \cos(x)} \)[/tex] is [tex]\( R - (4n+1)\frac{\pi}{4} \)[/tex].

Answer: (b) [tex]\( R - (4n+1)\frac{\pi}{4} \)[/tex]

### (iii) Range of [tex]\( f(x) = \sqrt{5 + 4\cos(x)} \)[/tex]

To find the range of this function, we need to look at the expression inside the square root, which is [tex]\( 5 + 4\cos(x) \)[/tex]:
- The cosine function [tex]\(\cos(x)\)[/tex] has a range of [tex]\([-1, 1]\)[/tex].
- Therefore, [tex]\( 4\cos(x) \)[/tex] ranges from [tex]\(-4\)[/tex] to [tex]\(4\)[/tex].
- Adding 5 to this range, we get [tex]\( 5 + 4\cos(x) \)[/tex], which ranges from [tex]\( 5 - 4 = 1 \)[/tex] to [tex]\( 5 + 4 = 9 \)[/tex].
- Taking the square root of these values, we get the range of [tex]\( \sqrt{5 + 4\cos(x)} \)[/tex] as [tex]\([ \sqrt{1}, \sqrt{9} ] = [1, 3]\)[/tex].

Answer: The range of [tex]\( f(x) = \sqrt{5 + 4\cos(x)} \)[/tex] is (1, 3)