restrictions
- Thread starter mathfun
- Start date
The question didn't ask for the restrictions of sin(x), as such; "determine the restrictions" is probably best read "Determine which values of x result in the expression being undefined".
Think of the annoying examiner who has to act like he or she doesn't know anything except what's written in the answer schedule: the examiner doesn't know what \(\displaystyle \sin{x} \, \neq \, 0, \, - 1\) means because the answers are in terms of x, so your best bet is to take it all the way and say \(\displaystyle x \, \neq \, 0, \, \pi, \, \frac{3\pi}{2}\) (for \(\displaystyle 0 \, \leq \, x \, \leq \2\pi\)) and get your deserved marks.
Think of the annoying examiner who has to act like he or she doesn't know anything except what's written in the answer schedule: the examiner doesn't know what \(\displaystyle \sin{x} \, \neq \, 0, \, - 1\) means because the answers are in terms of x, so your best bet is to take it all the way and say \(\displaystyle x \, \neq \, 0, \, \pi, \, \frac{3\pi}{2}\) (for \(\displaystyle 0 \, \leq \, x \, \leq \2\pi\)) and get your deserved marks.
Hello, mathfun!
It's simpler than you think . . .
\(\displaystyle \;\;\cot x\) is undefined for \(\displaystyle x\,=\,0,\,\pi.\;\;\)Hence: \(\displaystyle \,x\,\neq\,0,\,\pi\)
\(\displaystyle \;\;\)Since the denominator cannot be zero: \(\displaystyle \,\sin x\,\neq\,-1\;\;\Rightarrow\;\;x\,\neq\,\frac{3\pi}{2}\)
It's simpler than you think . . .
Restricting the domain to: \(\displaystyle 0\,\leq\,x\,<\,2\pi\) . . .
\(\displaystyle \;\;\cot x\) is undefined for \(\displaystyle x\,=\,0,\,\pi.\;\;\)Hence: \(\displaystyle \,x\,\neq\,0,\,\pi\)
\(\displaystyle \;\;\)Since the denominator cannot be zero: \(\displaystyle \,\sin x\,\neq\,-1\;\;\Rightarrow\;\;x\,\neq\,\frac{3\pi}{2}\)
You've demonstrated that you know cot(x) = cos(x)/sin(x), so is undefined when sin(x) = 0. And sin(x) = 0 when x=0, pi, in the domain 0<x<2pi. This has been covered.mathfun said:
A thread or two ago you appeared to demonstrate you know that sin(x) = -1 when x=3pi/2 in the domain 0<x<2pi.mathfun said:
-pi/2 and 3pi/2 are 2pi apart so are equivalent.
...0 being a restriction..what does that have anything to do with pi?Unco said:
A thread or two ago you appeared to demonstrate you know that sin(x) = -1 when x=3pi/2 in the domain 0<x<2pi.mathfun said:
-pi/2 and 3pi/2 are 2pi apart so are equivalent.[/quote]
then how come -pi/2 is not part of the restrictions?
