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

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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).
 
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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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