Trignometric functions and identities

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SUMMARY

The discussion focuses on solving maximum and minimum value problems for trigonometric functions using calculus, specifically the function f(x) = 10cos²(x) - 6sin(x)cos(x) + 2sin²(x). Participants suggest taking the derivative f'(x) = -20cos(x)sin(x) - 6cos²(x) + 6sin²(x) + 4sin(x)cos(x) and setting it to zero to find critical points. While this method is valid, it is noted that it may not be quicker than completing the square for the original function. The conversation also addresses the need to determine minimum values after finding maximums.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Knowledge of calculus, specifically differentiation
  • Familiarity with critical points and their significance in optimization
  • Ability to manipulate and simplify trigonometric expressions
NEXT STEPS
  • Learn how to apply the first derivative test for identifying maxima and minima
  • Study the method of completing the square for quadratic trigonometric expressions
  • Explore the use of second derivatives to confirm the nature of critical points
  • Investigate graphical methods for visualizing trigonometric function behavior
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in optimizing trigonometric functions using calculus techniques.

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Homework Statement


How to quickly solve problems on maximum and minimum values of trig functions with help of calculus:
Ex. 10cos2x-6sinxcosx+2sin2x

Homework Equations


none

The Attempt at a Solution


I know the method of simplification. But i want to do it quickly with calculus. How to do that??
 
Last edited:
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If you "want to do it quickly with calculus" then take the derivative, set the derivative equal to 0, and solve for x. However, the derivative is fairly complicated and I'm not sure this is "quicker" than just completing the square in the original.

Setting f(x)= 10cos^2(x)- 6sin(x)cos(x)+ 2sin^2(x) then f'(x)= -20cos(x)sin(x)- 6cos^2(x)+ 6sin^2(x)+ 4sin(x)cos(x)= -16sin(x)cos(x)- 6(sin^2(x)- cos^2(x))= 0.
 
OK. but we get maximum values from that what about minimum values?
 

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