Sketching A Graph Using Differentiation

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SUMMARY

The discussion focuses on the process of sketching a graph using differentiation, specifically addressing a step in the manipulation of trigonometric expressions involving cosine and sine functions. The key confusion arises in the transition from step 4 to step 5, where the participant questions the factoring of cos x and the application of the coefficient 1/6. The correct interpretation involves recognizing how to distribute the factor correctly within the brackets, leading to the expression cos x [ 1 - (1/6)(1 - 2sin² x - 2sin² x) ]. This highlights the importance of careful algebraic manipulation in calculus.

PREREQUISITES
  • Understanding of trigonometric identities, specifically sine and cosine functions.
  • Familiarity with differentiation techniques in calculus.
  • Knowledge of algebraic manipulation and factoring.
  • Basic understanding of graph sketching principles in calculus.
NEXT STEPS
  • Review trigonometric identities and their applications in calculus.
  • Practice differentiation of trigonometric functions using various techniques.
  • Explore algebraic manipulation strategies for simplifying expressions.
  • Learn about graph sketching techniques for functions involving trigonometric components.
USEFUL FOR

Students studying calculus, particularly those focusing on trigonometric functions and graph sketching, as well as educators seeking to clarify differentiation techniques and algebraic manipulation in their teaching.

themadhatter1
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Homework Statement



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Homework Equations





The Attempt at a Solution



I don't understand how they get from step 4 to step 5. Wouldn't you factor a cos x out of the brackets then have [tex](\cos x)(1-\frac{1}{6})[/tex] to the left of the brackets. Then you can multiply the [tex](1-\frac{1}{6})[/tex] inside the brackets.However that would yield [tex][\frac{5}{6}+\frac{10}{3}\sin^2x][/tex] I don't understand how they are getting 1- (1/6) times the stuff inside the brackets.
 

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[tex]\cos x - \frac{1}{6}[(1-2\sin^2 x)\cos x - 2\sin^2 x\cos x][/tex]

[tex]= \cos x - \frac{1}{6}[ \cos x (1-2\sin^2 x - 2\sin^2 x) ][/tex]

[tex]= \cos x - \frac{1}{6} \cos x(1-2\sin^2 x - 2\sin^2 x)[/tex]

[tex]= \cos x [ 1 - \frac{1}{6}(1-2\sin^2 x - 2\sin^2 x) ][/tex]
 

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