Answer :
Let's analyze the domain of the function [tex]\( y = \csc(x) \)[/tex].
The cosecant function is defined as:
[tex]\[ \csc(x) = \frac{1}{\sin(x)} \][/tex]
For [tex]\(\csc(x)\)[/tex] to be defined, [tex]\(\sin(x)\)[/tex] must not be zero because division by zero is undefined. Therefore, we need to determine where [tex]\(\sin(x) = 0\)[/tex].
The sine function, [tex]\(\sin(x)\)[/tex], is equal to zero at integer multiples of [tex]\(\pi\)[/tex]:
[tex]\[ \sin(x) = 0 \quad \text{when} \quad x = n\pi \quad \text{for any integer } n \][/tex]
Hence, [tex]\(\csc(x)\)[/tex] is undefined at [tex]\( x = n\pi \)[/tex]. The domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq n\pi \)[/tex] for any integer [tex]\( n \)[/tex].
The given statement in the question is:
"The domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex]."
Let's examine what this means. The expression [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex] suggests that [tex]\(\csc(x)\)[/tex] is undefined at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex].
However, we need to check if [tex]\(\sin(x)\)[/tex] is zero at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex]:
[tex]\[ \sin\left(\frac{\pi}{2} + n\pi\right) \][/tex]
For integer values of [tex]\( n \)[/tex]:
- When [tex]\( n \)[/tex] is even, [tex]\( \sin\left(\frac{\pi}{2} + 2k\pi\right) = 1\)[/tex] or [tex]\( -1\)[/tex].
- When [tex]\( n \)[/tex] is odd, [tex]\( \sin\left(\frac{\pi}{2} + (2k+1)\pi\right) = -1 \)[/tex] or [tex]\( 1\)[/tex].
In both cases, [tex]\(\sin\left(\frac{\pi}{2} + n\pi\right) \)[/tex] is either [tex]\( 1 \)[/tex] or [tex]\( -1 \)[/tex], which are non-zero. Therefore, [tex]\(\csc(x)\)[/tex] is indeed defined at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex].
Thus, the given statement that [tex]\( y = \csc(x) \)[/tex] is undefined at [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex] is incorrect. The correct statement is that the domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq n\pi \)[/tex].
Therefore, the answer is:
B. False
The cosecant function is defined as:
[tex]\[ \csc(x) = \frac{1}{\sin(x)} \][/tex]
For [tex]\(\csc(x)\)[/tex] to be defined, [tex]\(\sin(x)\)[/tex] must not be zero because division by zero is undefined. Therefore, we need to determine where [tex]\(\sin(x) = 0\)[/tex].
The sine function, [tex]\(\sin(x)\)[/tex], is equal to zero at integer multiples of [tex]\(\pi\)[/tex]:
[tex]\[ \sin(x) = 0 \quad \text{when} \quad x = n\pi \quad \text{for any integer } n \][/tex]
Hence, [tex]\(\csc(x)\)[/tex] is undefined at [tex]\( x = n\pi \)[/tex]. The domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq n\pi \)[/tex] for any integer [tex]\( n \)[/tex].
The given statement in the question is:
"The domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex]."
Let's examine what this means. The expression [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex] suggests that [tex]\(\csc(x)\)[/tex] is undefined at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex].
However, we need to check if [tex]\(\sin(x)\)[/tex] is zero at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex]:
[tex]\[ \sin\left(\frac{\pi}{2} + n\pi\right) \][/tex]
For integer values of [tex]\( n \)[/tex]:
- When [tex]\( n \)[/tex] is even, [tex]\( \sin\left(\frac{\pi}{2} + 2k\pi\right) = 1\)[/tex] or [tex]\( -1\)[/tex].
- When [tex]\( n \)[/tex] is odd, [tex]\( \sin\left(\frac{\pi}{2} + (2k+1)\pi\right) = -1 \)[/tex] or [tex]\( 1\)[/tex].
In both cases, [tex]\(\sin\left(\frac{\pi}{2} + n\pi\right) \)[/tex] is either [tex]\( 1 \)[/tex] or [tex]\( -1 \)[/tex], which are non-zero. Therefore, [tex]\(\csc(x)\)[/tex] is indeed defined at [tex]\( x = \frac{\pi}{2} + n\pi \)[/tex].
Thus, the given statement that [tex]\( y = \csc(x) \)[/tex] is undefined at [tex]\( x \neq \frac{\pi}{2} + n\pi \)[/tex] is incorrect. The correct statement is that the domain of [tex]\( y = \csc(x) \)[/tex] is [tex]\( x \neq n\pi \)[/tex].
Therefore, the answer is:
B. False
