1st year calc. trig, and inequalities

[m0286]

Hi!
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 I am 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 I've tried and tried I don't know how to solve this one. I've tried bringing x/9 to the other side and find a common denominator but that gives (9x^3-x)/9... and I don't 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 I am forgetting... I also tried mult. both sides by 9 9x^3<x.

 

[SpongeBobRhombusHat]

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.

SBRH

 

[Pseudo Statistic]

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

 

[Curious3141]

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.

$$x^3 \leq \frac{x}{9}$$

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

Now divide out by x. Separate into two cases.

Case 1 (x positive)

$$x^2 \leq \frac{1}{9}$$
$$0 < x \leq \frac{1}{3}$$
Adding the solution x = 0,
$$0 \leq x \leq \frac{1}{3}$$

Case 2 (x negative)
$$x^2 \geq \frac{1}{9}$$
$$x \leq -\frac{1}{3}$$

So the full solution set is $$(x \leq -\frac{1}{3}) \cup (0 \leq x \leq \frac{1}{3})$$

 

[HallsofIvy]

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 $\frac{1}{1+ cot^2 x}= \frac{1}{1+ \frac{cos^2 x}{sin^2 x}= \frac{sin^2 x}{sin^2 x+ cos^2 x}= sin^2 x$<br /> since sin² x+ cos² x= 1.<br /> <br /> 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.[/tex]

 

[arunbg]

Pseudo's method is a good one, but you need to sketch a cubic for the last part

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 ?

 

[Curious3141]

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 ?

No, I must say I haven't. What's the wavy curve method (sounds like something a hairdresser might use )?

 

[Hootenanny]

Heard of the wavy curve method ?

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

~H

 

[arunbg]

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 learned it under some other name or no name perhaps.
Anyway,
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.
HAHAHA

 

[Curious3141]

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 learned it under some other name or no name perhaps.
Anyway,
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.
HAHAHA

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

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.

 

[arunbg]

Sketching a curve or wavy curve its the same thing.
Just a little misunderstanding there.
And yeah I forgot about the coefficients.

Cheers,
Arun

 

[maverick280857]

Yes it is!

Suppose you have in general a cubic function $f(x)$, you can sketch the curve to get a fair idea of the location of roots (using $f'(x)$ 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 $x^2 < 9$ (this is too easy), you can write it as

$(x-3)(x+3) < 0$

draw a rough number axis divided into 3 parts: $x < -3$, $-3<x<3$ and $x > 3$. Write down the sign of the function $x^2 - 9 = (x-3)(x+3)$ in each of these regions. It is clear that the inequality is satisfied for $-3<x<3$. 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...