How to use trigonometric identities to solve for f'(x) = 0 in calculus 1

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SUMMARY

The discussion focuses on solving the calculus problem where the derivative f'(x) = cosX - 1/6cos3X is set to zero. The critical points occur at x = π/2 and x = 3π/2, where cosX equals zero. To further simplify the expression, the instructor suggests using trigonometric identities, leading to the form cosX[5/6 + 2/3sin²x]. The participant expresses confusion about the necessity of rewriting the derivative and the application of identities in this context.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with trigonometric functions and their properties.
  • Knowledge of trigonometric identities, including the addition formulas.
  • Ability to manipulate algebraic expressions involving trigonometric functions.
NEXT STEPS
  • Study the application of trigonometric identities in calculus problems.
  • Learn how to use the addition formula for cosine, specifically cos(3x) = cos(2x + x).
  • Explore the implications of critical points in the context of function analysis.
  • Practice rewriting derivatives using trigonometric identities to simplify expressions.
USEFUL FOR

Students in calculus courses, particularly those struggling with trigonometric identities and their application in solving derivatives. This discussion is also beneficial for tutors and educators looking to clarify these concepts for their students.

Whalstib
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Hi,

This is just a part of a calculus 1 problem. It occurs while attempting to make f'(x) = 0

f'(x) = cosX - 1/6cos3X

Now to make this zero I quickly note that in the domain of (0-2pi) cos=0 at pi/2 and 3pi/2 so the expression equals zero already! I have been informed by my instructor I need to use a series of identities to achieve a final answer close to

cosX[5/6 +2/3sin^2x]

While it may very well include the cos zero I outlined above (and in fact insists on it!) it is now more complicated... sigh...

Thing is I would never make this connection on exam day without seeing what is going on. Tutors give me that look like it's obvious as does my instructor but I just don't see how or why I should stretch it and why if so stop at this point as I could continue...

ANY help advise greatly appreciated!

Warren
 
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Whalstib said:
Hi,

This is just a part of a calculus 1 problem. It occurs while attempting to make f'(x) = 0

f'(x) = cosX - 1/6cos3X

Now to make this zero I quickly note that in the domain of (0-2pi) cos=0 at pi/2 and 3pi/2 so the expression equals zero already!
More precisely, the derivative is zero at those two values.
Whalstib said:
I have been informed by my instructor I need to use a series of identities to achieve a final answer close to

cosX[5/6 +2/3sin^2x]
I'm confused. Are you supposed to rewrite your derivative as above? If so, start by replacing cos(3x) with cos(2x + x) and using the addition formula.
Whalstib said:
While it may very well include the cos zero I outlined above (and in fact insists on it!) it is now more complicated... sigh...

Thing is I would never make this connection on exam day without seeing what is going on. Tutors give me that look like it's obvious as does my instructor but I just don't see how or why I should stretch it and why if so stop at this point as I could continue...

ANY help advise greatly appreciated!

Warren
 

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