Need Help with Calculus Summer Assignment on Cosine Identities?

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The discussion revolves around solving calculus-related assignments involving cosine identities. The user seeks help with finding cos(t) given cos(2t) = 1/3, and receives guidance on using the identity cos(2α) = 2cos²(α) - 1 to derive cos²(α) = 2/3. The conversation shifts to another problem involving the equation x² + y² + 3x = 0, where the user is unsure how to apply trigonometric identities. Participants clarify the use of the Pythagorean identity and the correct distribution of terms in the equation. The thread emphasizes the importance of understanding trigonometric identities in solving calculus problems.
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I'm taking Calculus next year and over the summer I have some assignments.
This one is due in a couple of hours, so any help would be appreciated!

If cos2t=1/3 and *0<_ 2t <_ pie, find cost. t=theta *less than or equal to

I don't know how to use the identities to help me.
:confused:
Please help!
 
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Well...

\cos{(2\alpha)} = 2\cos^2{(\alpha)}-1

Plug that in and post when you make progress.
 
ohh, costheta=sqrt(6)/3
?
 
2\cos^2{(\alpha)}-1 = \frac{1}{3}

2\cos^2{(\alpha)}=\frac{4}{3}

\cos^2{(\alpha)}=\frac{2}{3}

Can you finish from here?
 
oh yes thanks!

how about this one.

I'm not sure how to simplify it down, and how to distribute the ^2 once it has been plugged in.
x^2 + y^2 +3x=0 when x=rcostheta and y=rsintheta
 
Remember that \sin^2 x + \cos^2 x = 1. These questions don't seem to have anything to do with calculus, they just seem to be trigonometry.
 
(r\cos{\theta})^2 + (r\sin{\theta})^2+3(r\cos{\theta})=0

r^2\cos^2{(\theta)}+r^2\sin^2{(\theta)}+3(r\cos{\theta})=0

Do you see the trig identity coming in?
 
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the Pythag. Identity? Would you have to plug in rcostheta with the 3x?
 
I should have plugged that in earlier. But no, that's not where the identity comes in.

I'll give you my last hint to this problem.

r^2\cos^2{(\theta)}+r^2\sin^2{(\theta)}+3(r\cos{\theta})=0

r^2(\cos^2{\theta}+\sin^2{\theta})...
 
  • #10
OH! thanks!
 
  • #11
when you distribute the 3, would it be 3rcos3theta? or do you just not distribute the 3 to the cos?
 
  • #12
3r\cos{(\theta)}
 
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