Can Teaching Improve Your Learning Skill?

  • Thread starter Tyrion101
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In summary, the conversation discusses whether the first equation is a squared version of the second equation, and clarifies the use of trigonometric functions in the equations. It is also mentioned that the person asking the question may have some form of dyslexia, but it is not confirmed. Helpful tips are given on how to avoid misreading problems and to improve learning skills.
  • #1
Tyrion101
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Is the first a squared version of the other? I understand the trig function involved if it's just (2x) and tan is not squared.
 
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  • #2
No, I think what you're seeing is for twice the angle

Tan(2*theta)=tan(theta+theta)
 
  • #3
Tan squared is in general not equal to tan of twice the angle.
 
  • #4
But is it equal to: (2tanx/1-tan^2x)^2 is what I'm asking. I may have been unclear.
 
  • #5
Tyrion101 said:
But is it equal to: (2tanx/1-tan^2x)^2 is what I'm asking. I may have been unclear.
Use precise pure-text symbolism to remove all ambiguity. Using TEX would be better.
 
  • #6
(2tanx/1-tan^2x)^2
Trying TEX on my own, might work badly:
[tex](\frac{2tan(x)}{1-tan^2(x)})^2 [/tex]

I cannot think of any clever identity. Try breaking into sines and cosines.
 
  • #7
Tyrion101 said:
Is the first a squared version of the other? I understand the trig function involved if it's just (2x) and tan is not squared.
Now I understand what you asked.
No, those two are generally not equal.
 
  • #8
Ok thanks! I'm beginning to understand. I'm going to be glad when the trig identities homework is fini(finished in French.)
 
  • #9
Tyrion101 said:
Is the first a squared version of the other? I understand the trig function involved if it's just (2x) and tan is not squared.
@Tyrion101, despite what others have said in this thread, yes, ##\tan^2(2x)## is the square of ##\tan(2x)##.

Tyrion101 said:
But is it equal to: (2tanx/1-tan^2x)^2 is what I'm asking. I may have been unclear.
Yes and no. ##\tan^2(2x)## means ##[\tan(2x)]^2##, which in turn is equal to ## [\frac{2 \tan(x)}{1 - \tan^2(x)} ]^2##

In what you wrote, you are missing parentheses around the quantity in the denominator, 1 - tan2(x). What you wrote is the same as ##\frac{2\tan(x)}{1} - \tan^2(x)##
 
  • #10
Mark44 said:
@Tyrion101, despite what others have said in this thread, yes, ##\tan^2(2x)## is the square of ##\tan(2x)##.Yes and no. ##\tan^2(2x)## means ##[\tan(2x)]^2##, which in turn is equal to ## [\frac{2 \tan(x)}{1 - \tan^2(x)} ]^2##

In what you wrote, you are missing parentheses around the quantity in the denominator, 1 - tan2(x). What you wrote is the same as ##\frac{2\tan(x)}{1} - \tan^2(x)##
... and should be clear by now.
It be nice to know your history about your studies of Trigonometry, Tyrion101. Nothing wrong in studying Algebra 2 and Trigonometry more than once each, and one does not need to enroll in a class to do that. One should expect to learn material better through repeated study.
 
  • #11
Sorry for not completing the formula but I got my question answered. I get confused easily when frustrated or nervous, and in this case it was a bit of frustration. My two biggest problems in any math class seem to be the slowness at which I arrive at an answer, usually brought about by the other problem I have of misreading the question to begin with, I've long suspected that I have some form of dyslexia. For instance I saw a problem with -5beta (the actual symbol) and misread it to mean: -5/beta. Is there a at I can minimize this from happening?
 
  • #12
I'm not an expert on dyslexia or related problems, but if you misread 5β as 5/β, maybe you need glasses or contacts. That doesn't sound like dyslexia to me.

Also, before you get started working a problem, go back over the problem description to make sure that your first reading of it was correct.
 
  • #14
Mark44 said:
I'm not an expert on dyslexia or related problems, but if you misread 5β as 5/β, maybe you need glasses or contacts. That doesn't sound like dyslexia to me.

Also, before you get started working a problem, go back over the problem description to make sure that your first reading of it was correct.
It's other things too, like I'll do a problem 20 times only to find out that there is no sine2x at all. It's very frustrating. From what I understand from having gone to a school that helped dyslexic and add or adhd kids, it seems somewhat similar, but as you say Mark it might not be dyslexia related at all. Thanks for the link on math dyslexia I'd never heard of it before.
 
  • #15
As Sherlock Holmes might say:

We look but we don't not see.

You need to develop the over mind idea where as you're doing the problem you are looking over and validating what you did as you do it.

You can imagine yourself talking to yourself as you're doing the problem repeating what you're writing like you're teaching yourself.

Initially it may be hard to do and so it will take some practice to master.

Do you have someone you can teach to? By teaching you improve your learning skill.
 

1. What is the difference between Tan^2(2x) and Tan(2x)?

The main difference between Tan^2(2x) and Tan(2x) is that Tan^2(2x) represents the square of the tangent of 2x, while Tan(2x) represents the tangent of 2x. In other words, Tan^2(2x) is equal to [Tan(2x)]^2.

2. How do you simplify Tan^2(2x)?

To simplify Tan^2(2x), we can use the trigonometric identity Tan^2(x) = Sec^2(x) - 1. Therefore, Tan^2(2x) can be simplified to Sec^2(2x) - 1.

3. How do you graph Tan(2x)?

To graph Tan(2x), you can use a graphing calculator or plot points on a coordinate plane. The graph of Tan(2x) has a period of π and asymptotes at x = π/2 and x = -π/2. It also has a range of all real numbers.

4. What is the domain of Tan(2x)?

The domain of Tan(2x) is all real numbers except for values that make the denominator 0. In other words, the domain of Tan(2x) is all real numbers except for x = π/2 + πn, where n is any integer.

5. How do you solve equations involving Tan(2x)?

To solve equations involving Tan(2x), you can use the inverse tangent function, Arctan(x). Use the trigonometric identity Tan(2x) = Sin(2x)/Cos(2x) to rewrite the equation in terms of Sin(2x) and Cos(2x). Then, take the inverse tangent of both sides to isolate x.

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