Parametrics tangent line with point given

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SUMMARY

The discussion centers on simplifying the derivative dy/dx for a parametric equation involving trigonometric identities. The user presents their derivative as 4sin(θ)cos(θ) - 2csc²(θ) and seeks confirmation on whether it can be simplified further. The final expression they arrive at is -2sin³(θ)cos(θ), indicating a successful simplification process. The focus is on ensuring clarity in the simplification of trigonometric derivatives.

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  • Understanding of parametric equations
  • Knowledge of trigonometric identities
  • Familiarity with derivatives in calculus
  • Ability to manipulate algebraic expressions
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  • Study trigonometric identities and their applications in calculus
  • Learn techniques for simplifying derivatives of parametric equations
  • Explore the use of implicit differentiation in parametric contexts
  • Practice solving problems involving parametric equations and their slopes
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Students studying calculus, particularly those focusing on parametric equations and trigonometric derivatives, as well as educators looking for examples of derivative simplification techniques.

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Homework Statement


I understand the setup for finding the slope, but always get confused whether I've fully simplified when trig identities get involved. [/B]

Homework Equations


My dy/dx is [/B]
4sin(θ)cos(θ)
-2csc2(θ) which I simplified to just (-2sin(θ)cos(θ))/(csc2(θ)

Does that simpify to anything better than the way I have done it?

The Attempt at a Solution


Once I simplify the above expression fully, I know how to plug in the point to find the equation. Thanks so much!
 
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##-2\sin^3\theta \cos\theta##
 

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