CosA=cos2A+cos4A where 0°<or equal to A<or equal to 360°

  • Context: Undergrad 
  • Thread starter Thread starter johnibr5
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around solving the equation cosA = cos2A + cos4A for angles A within the range of 0° to 360°. Participants explore different methods and identities related to trigonometric functions as they attempt to find a solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests assistance with the equation, expressing difficulty in finding a solution despite prior experience.
  • Another participant suggests using the identity cos2A = 2cos²A - 1 and applying it to cos4A in a two-step process to form a polynomial in cosA.
  • A participant acknowledges a mistake in their approach and seeks clarification on where they went wrong.
  • Some participants note the importance of including cos2A in the calculations and recommend the two-step method as a simpler approach.
  • One participant mentions that their calculus course book presents a different method for solving the problem, leading to uncertainty about the two-step method's validity.
  • Another participant expresses confusion about the two-step method and requests further explanation.
  • Several participants share their progress, with one indicating they have resolved their issues and are now prepared for an exam.
  • Another participant mentions trying a different identity related to cosine sums but encountered difficulties near the end of their calculations.

Areas of Agreement / Disagreement

Participants express differing opinions on the best method to solve the equation, with some supporting the two-step method while others reference alternative approaches from their course materials. The discussion remains unresolved regarding the most effective solution strategy.

Contextual Notes

Some participants express uncertainty about specific steps in their calculations and the application of trigonometric identities, indicating potential gaps in understanding or execution. There is also mention of varying methods presented in different educational resources.

johnibr5
Messages
6
Reaction score
0
Hey guys, I'm doing a correspondence course in calculus and I have literally one last question to complete and I am having the worst luck solving it. I am coming here to ask because I hear the best minds are on this website.

Anyways the question, is:
Find the solution to the nearest angle for the following question

cosA=cos2A+cos4A where 0°<or equal to A<or equal to 360°

I'm usually decent in these types of questions but neither me or my aunt(really good in calc) can't figure it out.

Thanks and this really appreciated
 
Physics news on Phys.org
Use identity: cos2A = 2cos2A - 1.
Apply it also to cos4A in two steps (cos4A -> cos2A -> cosA).
You now have a polynomial in cosA.
Solve, keeping roots within interval [-1,1]. Find A (cos table if needed).
 
Last edited:
ImageUploadedByPhysics Forums1371761922.824723.jpg
thanks for the help, but I messed up somewheres I'm pretty sure. Could you tell me where I messed up? Thanks a bunch
 
hi johnibr5! :smile:

you've completely missed out the cos2A ! :redface:

also, it would be a lot easier to use mathman's :smile: two-step method …
mathman said:
Apply it also to cos4A in two steps (cos4A -> cos2A -> cosA).
 
Wow I feel dumb for forgetting the cos2A..
 
My calculus course book shows how to solve it differently so I'm not too sure about the 2 step method :p
 
johnibr5 said:
My calculus course book shows how to solve it differently so I'm not too sure about the 2 step method :p

your method would work fine if you hadn't materialised an extra cosA from nowhere! :rolleyes:

(the 2 step method is easier because it completely avoids any sines … try it! :smile:)
 
(just got up :zzz:)
johnibr5 said:
… Could you please explain that two-step method that mathman was talking about? I don't understand it yet..thanks a lot!

cos2A = 2cos2A - 1

so cos4A = 2cos22A - 1

= 2(2cos2A - 1)2 - 1​

now expand :smile:
 
Thank you so much! I finally got it, I knew I'd miss a little step throughout the way. Thanks again and now I can write this exam and enjoy comp sci!
 
  • #10
are you familiar with this identity ?
cos A + cos B = 2cos(A+B/2)*cos(A-B/2)
 
  • #11
Yeah, I tried that but messed up near the end. But I figured it out now, thanks
 

Similar threads

Undergrad Differentials of order 2 or bigger that are equal to 0
  • · Replies 5 ·
Replies
5
Views
2K
Coefficient of Restitution for a Hypothetical Collision where masses are not equal after the collision compared to before the collision
  • · Replies 6 ·
Replies
6
Views
1K
High School Why does the Limit Process give an Exact Slope?
  • · Replies 53 ·
2
Replies
53
Views
6K
High School Is 0.999... Equal to 0: A Discussion on Convergence and Infinite Series?
  • · Replies 44 ·
2
Replies
44
Views
5K
Undergrad Explanation and clarifications in epsilon-delta limit proofs
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Derivatives of Standard Functions
  • · Replies 17 ·
Replies
17
Views
2K
MHB Proof of the Equality of Supremums (Or Something Like That Anyway :) )
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Understanding the Concept of 'Cancelling dt's' in Derivatives and Integrals
  • · Replies 5 ·
Replies
5
Views
2K
Graduate Does the left side of Euler's Equation always equal zero?
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Measurement of heat as an interval measurement or a ratio measurement
  • · Replies 9 ·
Replies
9
Views
4K