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Homework Help: Trig Derivative

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Set f(x) = cos^2(x) + sin(x)
    Find the numbers x between 0 and 2pi where the tangent line to the graph of f is horizontal.

    2. Relevant equations
    pi/6 pi/2 3pi/2 11pi/6 = x

    I do not know how though

    3. The attempt at a solution

    -2sinx + cosx = 0
    Equivalent to zero because derivative = slope = 0 because horizontal

    Perhaps I have done it wrong lol, but I have no clue what to do from here if I did do it correct,
    Last edited: Jun 10, 2009
  2. jcsd
  3. Jun 10, 2009 #2
    Your problem is that the derivative of cos2x is not 2*sinx. remember the chain rule...
  4. Jun 10, 2009 #3
    So it is sin^2(x) = cos(x)= 0

    then -cos^2(x) +cos(x) +1 = 0


    2cos * -sin + cosx

    -2(sin)(cos) + cosx
    -2sin - cos = 0

    Lol uh oh ; (
  5. Jun 10, 2009 #4


    Staff: Mentor

    It's very difficult to follow what you're doing, particularly when you start with an equation, and then come up with some random appearing expression whose value you don't show.

    How did you get sin^2(x) = cos(x) = 0?

    Start with f(x) = cos^2(x) + sin(x)
    Find f'(x). (I.e., f'(x) = ...)
    Set f'(x) = 0.
    Solve for x.
  6. Jun 10, 2009 #5


    User Avatar
    Homework Helper

    That's the idea...
  7. Jun 11, 2009 #6
    cos^2(x) +sin(x)

    f' = -2(cos(sin) + cosx = 0

    (cosx(-2(sinx) +1) = 0

    cosx = 0

    pi/2, 3pi2

    sinx= 1/2

    pi/6, 11pi/6

    Is this done correctly?
  8. Jun 11, 2009 #7


    User Avatar
    Homework Helper

    Almost, but try to write mathematical expressions down correctly. Putting the brackets in the wrong places and forgetting arguments of functions will cause you to make mistakes and people that try to help you a headache.

    For example cos(sin) means nothing, but what do you mean with it? cos(sin(x)), or cos(x)sin(x)? I know that in this case you mean cos(x)sin(x), but the meaning of something should not be guessed in mathematics.

    You have [itex]f'(x)=-2 \cos x \sin x+\cos x=0[/itex].

    As for your answers, the cosine part is correct. The sine part is not, pi/6 is correct, but 11/pi/6 lies in the fourth quadrant and is -1/2 instead. The other value where sin(x)=1/2 lies in the second quadrant. What is it?
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