Trigonometry, Prove the Identity and more

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Homework Help Overview

The discussion revolves around proving various trigonometric identities and equations, including identities involving logarithmic functions and mechanics-related equations. The subject area includes trigonometry and calculus, with participants exploring simplifications and assumptions in their approaches.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants express confusion regarding specific identities and simplifications, with some attempting to relate known identities to the problems at hand. Questions arise about assumptions made in the context of mechanics equations and the validity of certain steps in the simplification process.

Discussion Status

Some participants have offered guidance on specific identities and simplifications, while others are exploring different interpretations of the problems. There is an ongoing exchange of ideas, with no explicit consensus reached on all points, but productive discussions are taking place.

Contextual Notes

Participants note constraints such as the assumption that certain variables are equal in the mechanics problem, and the challenges posed by the complexity of the equations involved. There is also mention of the educational context, with some participants sharing their backgrounds and motivations for studying these topics.

  • #31
dextercioby said:
Unfortunately f_{1}'(x)=f_{2}'(x) \nRightarrow f_{1}(x)=f_{2}(x).

True, but it does mean that the function may only differ by a constant = ]

So you in fact can use the method, but at the end you must try a test value of x, say 0, and find the constant.
 
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  • #32
but that would require more number of steps lol.
 
  • #33
50. ln |secx + tanx| = -ln |secx - tanx|
dy/dx of both sides = sec x

and ln |sec0 + tan0| = -ln |sec0 - tan0|
0=0

hmm.. only three steps!
 
  • #34
new problem, Prove the Identity

\tan^{-1}x+cot^{-1}x=\frac{\pi}{2}

i used the sum and difference identity of tan to prove that both sides were undefined ...

\tan^{-1}x+cot^{-1}x=\frac{\pi}{2}

\tan(\tan^{-1}x+cot^{-1}x)=\tan(\frac{\pi}{2})

\frac{\tan(tan^{-1}x)+\tan(\cot^{-1}x)}{1-\tan(\tan^{-1}x)\tan(cot^{-1}x)}=\tan\frac{\pi}{2}

latex isn't working ...

\frac{x+1/x}\{1-1} = tan(UND)

UND = UND

help with my logic please!
 
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  • #35
you know tan(pi/2) = UD
but cot(pi/2) = 0

so, take cot of both sides
and you can write it as
1/tan(all side)
 
  • #36
rootX said:
you know tan(pi/2) = UD
but cot(pi/2) = 0

so, take cot of both sides
and you can write it as
1/tan(all side)
well wouldn't i need the sum and difference identity for cot?

i proofed it the way i got the sum/diff for tan and got 0 = 0

EDIT: crap, i messed up, let me re-work this ... my handwriting is very bad, lol

by luck i arrived at the same answer, 0 in my numerator.
 
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  • #37
rocophysics said:
well wouldn't i need the sum and difference identity for cot?

no, you don't
cot\left( atan\left( x\right) +acot\left( x\right) \right) =0
\frac{1}{tan\left( atan\left( x\right) +acot\left( x\right) \right) }=0
 
  • #38
\tan (\pi/2 -x) = \cot x.
 
  • #39
my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing
 
  • #40
he squared them, or used that assumption?
 
  • #41
rocophysics said:
my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing

I know I'm jumping into a thread I've probably not read properly here, but in question 52 it says that "I_1 can equal I_2 in some situations in mechanics. Assuming this is the case, simplify." Therefore, one simply sets I_1=I_2, which happens to simplify the equation a great deal.
 
  • #42
here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.
 
  • #43
1=2

nicktacik said:
Nope, otherwise you could prove that 1=2.

Which is clearly true.
 
  • #44
P.O.L.A.R said:
here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

after, spending an hour, I plugged in some random number:
f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)

f(6)
-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)

I guess that really doesn't give a 0!
 
  • #45
rootX said:
after, spending an hour, I plugged in some random number:
f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)

f(6)
-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)

I guess that really doesn't give a 0!

You have to remember that when working with proofs you can only work one side of the proof not both. Usually it is best to start with the side that has the most variables.
 
  • #46
If rootX is correct is that his expression doesn't not give zero, his *proof* is always a disproof by form of the counterexample n=6.
 
  • #47
P.O.L.A.R said:
here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

arctan(1/n) = arctan(1/[n+1])+arctan(1/[n^2+n+1])

forgot the n sorry. also I thought I put arctan but I guess not wow I really messed that one up I aologize. A little tired I guess.
 
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  • #48
Apply tangent on both sides of the equation. What do you get ?
 
  • #49
dextercioby said:
Apply tangent on both sides of the equation. What do you get ?

I already proved it myself people were asking proof questions and I found this one to be a good one.

Starting that way didnt work for me. I couldn't figure out where to go after taking the tan.
 
Last edited:

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