can someone please explain what principle domain is and why the answer is a not c?

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What is the principle domain of cosec(x)?

a.
R−(2nπ)

b.
R

c.
(−1,1)

d.
R−(nπ)

cosec(x) = 1/sin(x)
so the (principle?) domain of cosec is the "same" as sin, apart from where sin=0.

Not really sure what your multiple choice answers are due to formatting. They appear to be the usual domain, rather than the principle domain). c) corresponds to the range of sin().

The domain of the trig function is the set of values of x for which its defined. For sin and cos, it corresponds to all real values. So for cosec it would be all real values apart from integer multiples of pi which is when sin(x) = 0 so cosec(x) is undefined (just sketch the function). This looks like d) rather than a).

(edited 2 weeks ago)

Reply 2

Nitrotoluene

The domain of cosec(x) is the set of all real numbers except for the points where the sine function is zero.
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Reply 3

$Avatar for efsfsf\\eeeeeee$

efsfsf\\eeeeeee

Original post by mqb2766

cosec(x) = 1/sin(x)
so the (principle?) domain of cosec is the "same" as sin, apart from where sin=0.
Not really sure what your multiple choice answers are due to formatting. They appear to be the usual domain, rather than the principle domain). c) corresponds to the range of sin().
The domain of the trig function is the set of values of x for which its defined. For sin and cos, it corresponds to all real values. So for cosec it would be all real values apart from integer multiples of pi which is when sin(x) = 0 so cosec(x) is undefined (just sketch the function). This looks like d) rather than a).

thanks, but cant you still have for example pi - 0, if you let R = pi and n = 0, which would cause cosecx to be undefined but it would still be in the domain?

Reply 4

mqb2766

Original post by efsfsf\\eeeeeee

thanks, but cant you still have for example pi - 0, if you let R = pi and n = 0, which would cause cosecx to be undefined but it would still be in the domain?

cosec(x) is undefined at x=0, +/-pi, ... so those points must be excluded from the domain. The domain is the set of points which were allowed to plug into the function. But the function is not defined/undefined for those points, so they must be excluded from the domain.

The function 1/x has a domain R\0 as it is defined for all real x apart from x=0 where its undefined.

(edited 1 week ago)

Reply 5

davros

Original post by efsfsf\\eeeeeee

thanks, but cant you still have for example pi - 0, if you let R = pi and n = 0, which would cause cosecx to be undefined but it would still be in the domain?

Out of interest, what do you mean by "let R = pi"? R represents the set of all real numbers, so you can't assign a value to it :smile:

Reply 6

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efsfsf\\eeeeeee

Original post by davros

Out of interest, what do you mean by "let R = pi"? R represents the set of all real numbers, so you can't assign a value to it :smile:

Sorry I meant since R includes all real numbers, if you took pi as an example of that set and did that calculation wouldn't the problem occur?

Reply 7

davros

Original post by efsfsf\\eeeeeee

Sorry I meant since R includes all real numbers, if you took pi as an example of that set and did that calculation wouldn't the problem occur?

Indeed, which is why the correct answer looks like d) not a) as suggested above! Are you able to post an image of the question and mark scheme for this? Where does the question come from?

Reply 8

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efsfsf\\eeeeeee

Original post by mqb2766

cosec(x) is undefined at x=0, +/-pi, ... so those points must be excluded from the domain. The domain is the set of points which were allowed to plug into the function. But the function is not defined/undefined for those points, so they must be excluded from the domain.
The function 1/x has a domain R/0 as it is defined for all real x apart from x=0 where its undefined.

Ok and is R - 2npi wrong bc it allows for x=pi?

And also can you tell me in words what R - npi means, so I can understand better?

Reply 9

davros

Original post by efsfsf\\eeeeeee

Ok and is R - 2npi wrong bc it allows for x=pi?
And also can you tell me in words what R - npi means, so I can understand better?

Assuming the R is formatted in the usual way for sets, R - npi means "R without integer multiples of pi", R - 2npi means "R without the multiples of 2pi" or "R without the even multiples of pi". It doesn't exclude pi, 3pi etc, which is why (a) doesn't look right, but it would be helpful to see the original correctly formatted version of the question and mark scheme :smile:

(edited 1 week ago)

Reply 10

davros

Original post by Nitrotoluene

The statement is true, IMHO...of course!
The set of all real numbers, excluding multiples of 2pi, is denoted by R - 2pin. This means that all real numbers such as pi, 3pi, 4pi, etc. are included in the set, but numbers such as 2pi, 4pi, 6pi, etc. are not.