Solve Trig Asymptotes: Find Equation on -pi < x < pi

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The discussion focuses on finding the asymptotes of the function tan(2 sin x) within the interval -pi < x < pi. The initial approach involves setting cos(2 sin x) to zero, leading to two asymptotes derived from the equations 2 sin x = pi/2 and 2 sin x = -pi/2. However, the graph reveals four asymptotes, prompting a deeper analysis of the tangent function's behavior. It is noted that the argument of the tangent function can take on specific values, requiring an exploration of how often 2 sin x reaches these values within the given interval. The conversation concludes with a sense of clarity and appreciation for the problem-solving process.
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Homework Statement



Find the equation of the asymptotes.
On the interval -pi < x < pi

Homework Equations



tan(2 sin x)

The Attempt at a Solution



sin(2 sin x)/cos( 2 sin x)

Set cos( 2 sin x) = 0

2 sin x = arccos(0) = pi/2
2 sin x = arccos(0) = -pi/2

x = +-(arcsin pi/4)

But these are only but two asymptotes on the interval -pi < x < pi of tan( 2 sin x)-- yet when graphed; there are 4 observed.
 
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Use what you know of the tangent function directly.
tan has assymtotes when it's argument is a particular series of numbers - what are they?
2sin(x) can only evaluate to two of those numbers - what are they?
How many times does 2sin(x) visit each of those numbers in the interval?
 
Haha whilst formulating my response... I saw the light! Thank you so much!

Great way to start the day, and good one to ya!
 
Often the way - well done.
 
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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