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1st year calc. trig, and inequalities

  1. May 10, 2006 #1
    Im stuck on 2 assignment questions and I was hoping to get help on whut I am doing wrong. Its 1st year Calculus

    1) It says Given a right angled triangle prove that 1/1+cot^2 X=sin^2 X
    so I know cot=1/tan so 1/tan= 1/(opp/adj) therefore cot=1/(opp/adj) so 1/cot become (this is where it gets messy) opp/adj..?? i think since cot=1/(opp/adj) =adj/opp so 1/cot youd flip it again so opp/adj??? And sin is opp/hyp .. so Im confused how does 1+opp/adj=opp/hyp ... or is that just it and it doesnt??

    2) is *Solve the given inequalities giving the solution set of each s an interval or as a union of intervals. x^3 <x/9 so Ive tried and tried I dont know how to solve this one. ive tried bringing x/9 to the other side and find a common denominator but that gives (9x^3-x)/9... and I dont know how to get x on one side and the numbers onthe other... especially since there is an x and a x^3.... Is there a set of rules or something Im forgetting... I also tried mult. both sides by 9 9x^3<x.
  2. jcsd
  3. May 11, 2006 #2
    For 1) you should use the fact that cot(x)= cos(x) / sin(x) instead of the adjacent/opposite stuff. Then try to manipulate the given equation to look like an identity that you already know.

  4. May 11, 2006 #3
    x^3 < x/9
    Multiply both sides by 9 and move the term on the right:
    9x^3 - x < 0
    x(9x^2 - 1) < 0
    And factorizing further...
    x(3x + 1)(3x - 1) < 0
    You can go on from there. :)
  5. May 11, 2006 #4


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    Pseudo's method is a good one, but you need to sketch a cubic for the last part (not difficult in itself).

    Here's another method.

    [tex]x^3 \leq \frac{x}{9}[/tex]

    First check that x = 0 is a solution, it is.

    Now divide out by x. Separate into two cases.

    Case 1 (x positive)

    [tex]x^2 \leq \frac{1}{9}[/tex]
    [tex]0 < x \leq \frac{1}{3}[/tex]
    Adding the solution x = 0,
    [tex]0 \leq x \leq \frac{1}{3}[/tex]

    Case 2 (x negative)
    [tex]x^2 \geq \frac{1}{9}[/tex]
    [tex]x \leq -\frac{1}{3}[/tex]

    So the full solution set is [tex](x \leq -\frac{1}{3}) \cup (0 \leq x \leq \frac{1}{3})[/tex]
    Last edited: May 11, 2006
  6. May 11, 2006 #5


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    For the first problem, I would be inclined to change everything to sine and cosine.
    [tex]cot x= \frac{cos x}{sin x}[/itex] so [itex]\frac{1}{1+ cot^2 x}= \frac{1}{1+ \frac{cos^2 x}{sin^2 x}= \frac{sin^2 x}{sin^2 x+ cos^2 x}= sin^2 x[/itex]
    since sin2 x+ cos2 x= 1.

    I'm a bit puzzled by "Given a right angled triangle". That seems to me to imply that the equation is true only if x is an angle in a right triangle but in fact that is an identity, true for all x.
  7. May 11, 2006 #6
    Why do you mean by sketching a cubic?
    You can get the answer directly from Pseudo's last step with no calculations.
    Heard of the wavy curve method ?
  8. May 11, 2006 #7


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    No, I must say I haven't. What's the wavy curve method (sounds like something a hairdresser might use :rofl:)?
  9. May 11, 2006 #8


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    That's new to me. What is it? I'm intrigued now :smile:

  10. May 14, 2006 #9
    Well, its just a silly name given to a simple method of solving such equations in math (I guess it was coined by some Indian as I see it only in Indian texts).You have probably learnt it under some other name or no name perhaps.
    If you have a polynomial function like (ax+b)(cx+d)....... ,
    and would like to know where the function is +ve or -ve( on the real line of course) just arrange the roots in ascending order as on the no line.
    Start from the rightmost root.All values of the function for x greater than this root are +ve.Then in the interval between the largest and second largest the value is -ve and its goes on alternating between the roots.

    As an example , in the above case x(3x+1)(3x-1),
    The roots are -1/3,0,1/3.
    x > 1/3 => expression +ve
    0< x < 1/3 => -ve
    -1/3< x < 0 =>+ve
    x<-1/3 => -ve

    Sounds familiar ?

    Its called wavy curve because you can draw something of a wiggling worm on the no line.
  11. May 14, 2006 #10


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    But that's *exactly* the same as roughly sketching the curve to show the roots!:confused:

    Also, don't forget that this method only forms the correct conclusions when the coefficient of the highest power of x is positive. If it's negative, the rule should be reversed for everything.

    I'm sorry, this just doesn't strike me as anything to write home about, it's basically the same thing I do when I sketch the curve.:smile:
  12. May 14, 2006 #11
    Sketching a curve or wavy curve its the same thing.
    Just a little misunderstanding there.
    And yeah I forgot about the coefficients.

  13. May 14, 2006 #12
    Yes it is!!

    Suppose you have in general a cubic function [itex]f(x)[/itex], you can sketch the curve to get a fair idea of the location of roots (using [itex]f'(x)[/itex] which is a quadratic and easy to solve). You can also use negative Rolle's Theorem if its applicable.

    As a simple example, if you have something like [itex]x^2 < 9[/itex] (this is too easy), you can write it as

    [itex](x-3)(x+3) < 0[/itex]

    draw a rough number axis divided into 3 parts: [itex]x < -3[/itex], [itex]-3<x<3[/itex] and [itex]x > 3[/itex]. Write down the sign of the function [itex]x^2 - 9 = (x-3)(x+3)[/itex] in each of these regions. It is clear that the inequality is satisfied for [itex]-3<x<3[/itex]. You can try doing this for cubic functions factorizable as 3 linear functions or a quadratic with complex factors and a linear factor. More examples will help...
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