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

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  • Thread starter Thread starter johnibr5
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SUMMARY

The discussion centers on solving the equation cosA = cos2A + cos4A within the interval 0° ≤ A ≤ 360°. Participants recommend using the identity cos2A = 2cos²A - 1 and applying it in a two-step method to simplify cos4A into a polynomial in terms of cosA. The solution involves solving this polynomial while ensuring the roots remain within the interval [-1, 1]. The two-step method is highlighted as a more efficient approach to avoid complications with sine functions.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos2A and cos4A.
  • Familiarity with polynomial equations and their roots.
  • Basic knowledge of calculus concepts related to angle measures.
  • Ability to apply trigonometric tables for solving equations.
NEXT STEPS
  • Learn how to derive and apply trigonometric identities like cos2A = 2cos²A - 1.
  • Study polynomial equations in trigonometry and methods for solving them.
  • Explore the two-step method for simplifying trigonometric expressions.
  • Review the use of trigonometric tables for finding angle measures.
USEFUL FOR

Students in calculus, particularly those studying trigonometric equations, as well as educators and tutors looking for effective methods to teach these concepts.

johnibr5
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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
 
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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
 

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