Limiting Values of 2cos^2(x)/cos(3x) at x->pi/2

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (1 + cos(2x)) / cos(3x) as x approaches π/2. The subject area involves trigonometric limits and identities.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various trigonometric identities and limit properties, questioning the effectiveness of their approaches and suggesting alternative methods such as L'Hôpital's rule. There is discussion about potential errors in reasoning and the implications of using specific trigonometric substitutions.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on the limit evaluation process. Some have proposed specific substitutions and transformations, while others have pointed out potential flaws in reasoning. There is a recognition of the need to clarify certain steps and assumptions, but no consensus has been reached.

Contextual Notes

Participants are working under the constraints of standard trigonometric identities and limits, and there is an acknowledgment of potential arithmetic errors in the calculations presented.

Dell
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lim
x->pi/2 (1+cos2x)/cos3x

i tried all the identities and got

2cos^2(x)/cos(3x)

what now? i used cos(a+b)=cos(a)*cos(b)-sin(a)sin(b)
but no help there
 
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I don't think you tried 'all of the identities'. The denominator hasn't changed at all. What can you do to it?
 
the best i could come up with is

lim
x->pi/2 (1+cos2x)/cos3x

lim
x->0 2cos^2(pi/2-x)/cos(3(pi/2-2))

lim
x->0 (2sin^2(x))/sin(3x)

lim
x->0 2sin(x)*-1/3

=0
 
Last edited:
can you use l'hospital's? good to check with anyway...
 
EDIT: as below

in the mean time its worth trying Dick's suggestion though...
 
Last edited:
Dell said:
the best i could come up with is

lim
x->pi/2 (1+cos2x)/cos3x

lim
x->0 2cos^2(pi/2-x)/cos(3(pi/2-2))

lim
x->0 (2sin^2(x))/sin(3x)

lim
x->0 2sin(x)*1/3

=0

Ok, you can do it that way. cos(3(pi/2-x))=(-sin(3x)), though.
 
lanedance said:
With last comment in mind, I'm trying to pick the flaw in you logic... i think you miss a negative when you subsitute x->pi/2-x, and can't just cancel the (2sin^2(x))/sin(3x) to 2sin(x)*1/3.

This is beacasue of the trig identities allow you to write sin^2x as sin2x and so on, think what thismeans in terms of the taylor series...

in the mean time its worth trying Dick's suggestion though...

I guess I was assuming the OP knew lim sin(x)/sin(3x)=(1/3) and didn't think sin(x)/sin(3x)=(1/3). Hope I wasn't wrong.
 
nope, my bad, made a quick arithmetic error, all looks good
 

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