Need Help with Calculus Summer Assignment on Cosine Identities?

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    Calculus Summer
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SUMMARY

The discussion focuses on solving calculus-related problems involving cosine identities. The key identity used is the double angle formula, \(\cos(2\alpha) = 2\cos^2(\alpha) - 1\), which helps derive \(\cos(\theta)\) when given \(\cos(2t) = \frac{1}{3}\). Participants also address a trigonometric equation involving \(x\) and \(y\) expressed in polar coordinates, leading to the application of the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). The conversation emphasizes the importance of correctly applying trigonometric identities in calculus problems.

PREREQUISITES
  • Understanding of trigonometric identities, specifically the double angle formula.
  • Familiarity with polar coordinates and their conversion to Cartesian coordinates.
  • Knowledge of the Pythagorean identity in trigonometry.
  • Basic calculus concepts, particularly in relation to trigonometric functions.
NEXT STEPS
  • Study the derivation and applications of the double angle formulas in trigonometry.
  • Learn how to convert between polar and Cartesian coordinates effectively.
  • Explore the Pythagorean identity and its implications in solving trigonometric equations.
  • Practice solving calculus problems that incorporate trigonometric identities and functions.
USEFUL FOR

Students preparing for calculus, particularly those needing assistance with trigonometric identities and their applications in calculus problems. This discussion is beneficial for anyone looking to strengthen their understanding of the relationship between trigonometry and calculus.

hotrocks007
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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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