- **Calculus**
(*http://mymathforum.com/calculus/*)

- - **stationary points of a trig function**
(*http://mymathforum.com/calculus/343130-stationary-points-trig-function.html*)

stationary points of a trig functionHi, am I doing this right? Find the coordinates when the gradient of this function is equal to zero t = cos(θ) f '(t) = -sin(θ) $\ \ \ \ \ \ \ \ $ = -sin(θ) = 0 $\ \ \ \ $ θ = 0/-sin $\ \ \ \ $ θ = 0 Plug 0 in to original equation t = cos(0) t = 1 Coordinates = 0,1 where gradient is zero. |

$\sin\theta$ is periodic. |

Having obtained -sin(θ) = 0, one cannot divide by "-sin". Instead, solve sin(θ) = 0 by using knowledge of the sin function. |

All times are GMT -8. The time now is 03:46 PM. |

Copyright © 2018 My Math Forum. All rights reserved.