Need help with determing domains of sin, cos, and tan

AI Thread Summary
The discussion focuses on understanding the domains of the trigonometric functions sine (sin), cosine (cos), tangent (tan), and their inverses. The domains for sin and cos are all real numbers, while tan is defined for all real numbers except odd multiples of π/2, which can be expressed as (2n + 1)π/2, where n is an integer. The cosecant function (csc) is also excluded at integer multiples of π, as it is the reciprocal of sin. Participants emphasize the importance of understanding why certain values are excluded, particularly in relation to division by zero. Engaging with interactive tools is suggested to visualize these concepts better.
name_ask17
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Homework Statement



Ok, this is not really a problem, but I need help on understanding the basics of sin, cos, tan, and their inverses.

i was looking at http://www.analyzemath.com/DomainRange/domain_range_functions.html and it was saying that the domain for sin and cos is (-inf , + inf)
and then for tan it is All real numbers
except pi/1 + n*Pi
but then for csc, it is All real numbers
except n*Pi

Can you explain why? I think I'm having trouble with figuring out how to find domains and I want to understand this before I start calculus. Please explain.
 
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You might want to check the expression for x where tan x goes to +inf or -inf.
 
You need to try a little yourself before asking, but I'll aid you this once..

Tan(x) = Sin(x)/Cos(x)

csc(x) = 1/Sin(x)

What's the rule for dividing?
 
name_ask17 said:

Homework Statement



Ok, this is not really a problem, but I need help on understanding the basics of sin, cos, tan, and their inverses.

i was looking at http://www.analyzemath.com/DomainRange/domain_range_functions.html and it was saying that the domain for sin and cos is (-inf , + inf)
and then for tan it is All real numbers
except pi/1 + n*Pi
I don't know what to make of "pi/1 + n*Pi ". The domain for the tangent function is all real numbers x, such that x ≠ (2n + 1)∏/2, where n is an integer. IOW, all reals except odd multiples of ∏/2.
 
name_ask17 said:
and then for tan it is All real numbers
except pi/1 + n*Pi
Correction: this should be
\frac{\pi}{2} + n\pi
. And you have to specify what n can equal, as Mark44 said.
 
Hi Name_Ask,

I'd like you to go to http://www.touchtrigonometry.org/" and play around with it a little bit.

While you're there, make sure to do the following:

  • Look at the bottom left of the screen where you see the tig. function names and a value beside each.
  • Turn them all off by clicking on them.
  • Turn one on at a time.
  • Take notice of how often its pattern repeats, and when it starts.
  • Examine all the "x" values it can hold and the ones that are impossible.
  • Why are some of these Tig values impossible?
  • Click the active graph at any time to "Pause" your mouse, and look at what the line does on the Trig Circle to the left.
  • Compare what you see with your knowledge of what happens when a number is divided by 0.
  • Repeat with a new trig function.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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