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

1. Jun 20, 2013

### johnibr5

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) cant figure it out.

Thanks and this really appreciated

2. Jun 20, 2013

### mathman

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: Jun 20, 2013
3. Jun 20, 2013

### johnibr5

thanks for the help, but I messed up somewheres I'm pretty sure. Could you tell me where I messed up? Thanks a bunch

4. Jun 20, 2013

### tiny-tim

hi johnibr5!

you've completely missed out the cos2A !

also, it would be a lot easier to use mathman's two-step method …

5. Jun 20, 2013

### johnibr5

Wow I feel dumb for forgetting the cos2A..

6. Jun 20, 2013

### johnibr5

My calculus course book shows how to solve it differently so I'm not too sure about the 2 step method :p

7. Jun 20, 2013

### tiny-tim

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

(the 2 step method is easier because it completely avoids any sines … try it! )

8. Jun 21, 2013

### tiny-tim

(just got up :zzz:)
cos2A = 2cos2A - 1

so cos4A = 2cos22A - 1

= 2(2cos2A - 1)2 - 1​

now expand

9. Jun 21, 2013

### johnibr5

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. Jun 21, 2013

### Intrastellar

are you familiar with this identity ?
cos A + cos B = 2cos(A+B/2)*cos(A-B/2)

11. Jun 21, 2013

### johnibr5

Yeah, I tried that but messed up near the end. But I figured it out now, thanks