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How to find derivative critical points that DON'T use ref. angles?

  1. Jul 22, 2013 #1
    1. The problem statement, all variables and given/known data
    I am trying to figure out how to solve for problems that give a function and ask for the critical points. However, the ones I am having difficulty with are the ones that have multiple trigonometric functions included in them.


    2. Relevant equations
    Find the critical points of the following function on the given interval:
    f(x) = 8sin(x)+3cos(x) ; [-2∏, 2∏]


    3. The attempt at a solution
    Once I find the derivative, I grouped the trigonometric functions onto one side to make:
    8/3 = tan(x)

    However, I am at a loss as to how I can solve for x here? Our instructor never explained how we can solve for trigonometric functions without using the reference angles. I am trying to brush up on this stuff since I have calc II this fall, but am finding a couple things like this that we didn't cover.

    Please keep in mind when providing your solution and method to that solution that I have only taken calc I up to this point. Just throwing that out there in hopes that you take this into consideration and don't throw things out there that might go over my head at this point of my education. Thank you.
     
    Last edited: Jul 22, 2013
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  3. Jul 22, 2013 #2

    Zondrina

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    ##tan^{-1}(tan(x)) = x##

    Use the inverse tangent ( arc tangent if you prefer denoted arctan(x) ).

    So also ##arctan(tan(x)) = x##
     
  4. Jul 22, 2013 #3

    verty

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    Draw a rough tan graph over the interval you are working with, notice where tan(x) > 0. Find a general formula for x using the ##tan^{-1}## function (tan is periodic with period π).

    The most important thing to remember with these is to be absolute certain about the range of the inverse function.
     
  5. Jul 22, 2013 #4

    vanhees71

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    You should be aware that the statement of Zendrina is highly misleading and very dangerous! The usual real-valued arctan function is defined as
    [tex]\arctan:\mathbb{R} \rightarrow (-\pi/2,\pi/2),[/tex]
    which is known as the main branch of the arctan function in complex function theory.

    So for a given number [itex]y \in \mathbb{R}[/itex] you find one specific angle [itex]x \in (-\pi/2,\pi/2)[/itex] such that [itex]\tan x=y[/itex]. Now, you want to find not only this one specific solution for you equation but all. It's however easy to find all these (infinitely many!) other solutions by thinking carefully about the properties of the tangens. It's best to draw the function [itex]y=\tan x[/itex] and also the function [itex]y=8/3[/itex] to see what's going on here!
     
  6. Jul 22, 2013 #5
    Thank you for the feedback. As far as the graphs of each and what is happening, that part is easy to me. However, the problem wants the exact values, which is where I am having a hard time with this. Is it something you would typically need a graphing program or calculator for, or is there a way to do it without one? Either way, I'm thinking now might be a good time for me to learn how to find graph intersections on my calc.
     
  7. Jul 22, 2013 #6

    Ray Vickson

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    You wrote the interval as [-2/∏, 2/∏]. Do you mean [-π/2,π/2] or [-2π,2π]?
     
  8. Jul 22, 2013 #7
  9. Jul 23, 2013 #8

    Ray Vickson

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    You can always reduce ##8\sin(x)+3\cos(x)## to ##a \sin(x+b)## for constants ##a,b## that are related to 8 and 3; this is a very standard trick/method, which you would be well advised to learn. Anyway, that reduces your problem to one involving a single trig. function.
     
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