Wednesday, May 20, 2015

Analysis Quiz 7: IMO Mock Trigonometric Math Quiz

Analysis Quiz 6:

IMO Mock Trigonometric Math Quiz

Please answer the following questions based on the trigonometric equation below:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

Question 1.

Would you convert the cosecant function to the sine function to solve the trigonometric equation above?



For an old hand like me (I'm not old at all, hehehe...), I could say right up front, loudly that I would not convert the cosecant function to the sine function in this case! But, if you are new and eager to learn, I will lead you to the answer, just that you have to be patient.

Let's say you don't know if that would be an appropriate move, you would try it out on a piece of paper end get:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

$\sin 9x-\dfrac{1}{\sin^2 x}=5\sin 3x+9\tan^2 x-1$

You might notice you could try to rewrite the function of tangent as well that gives:

$\sin 9x-\dfrac{1}{\sin^2 x}=5\sin 3x+9\left(\dfrac{\sin^2 x}{\cos^2 x}\right)-1$

The next thing that you would do without a question is to clear the fraction on the left by multiplying both sides of the equation by the quantity of $\sin^2 x \,\cos^2 x$:

$\sin 9x\sin^2 x\cos^2 x-\cos^2 x=5\sin 3x\sin^2 x\cos^2 x+9\sin^2 x \sin^2 x-\sin^2 x\cos^2 x$

$\sin 9x\sin^2 x\cos^2 x-\cos^2 x=5\sin 3x\sin^2 x\cos^2 x+9\sin^4 x -\sin^2 x\cos^2 x$

Things started to get complicated from this point, but we can still manage the equation:

$\sin 9x(\sin x\cos x)^2-\cos^2 x=5\sin 3x(\sin x\cos x)^2+9\sin^4 x -(\sin x\cos x)^2$

$(\sin x\cos x)^2(\sin 9x-5\sin 3x+1)=9\sin^4 x +\cos^2 x$

[MATH]\color{yellow}\bbox[5px,purple]{(\sin x\cos x)^2(\sin 9x-5\sin 3x+1)=9\sin^4 x +1-\sin^2 x}[/MATH]

Remember that we're asked to solve for this equation, and there is no way what we have obtained in the equation highlighted in purple above is going to help.

So the answer to this problem whether one should convert the cosecant function to sine function is a decidedly no.

Question 2:

Do you think the trigonometric identity $1+\cot^2x=\csc^2x$ would be useful in solving this problem?



Yes, actually question 1 and 2 are deeply connected. Since the convert from cosecant function to a sine function spells disaster, we would rather change it to cotangent function as it is related to the tangent function.

Therefore the answer for question 1 and 2 should be contrary to one another, that means we should get a Yes for this question.

Question 3:

Which of the following method(s) is/are possible to use to solve the given trigonometric equation?

Differentiation +Trigonometric Identities
AM-GM Inequality+Trigonometric Identities
Strictly Trigonometric Identities


This is the toughest part for this quiz because in this problem I gave you what might be the correct or incorrect representations which could use to solve the problem.

Yes, you learned differentiation in calculus before, so did you learn trigonometric identities and AM-GM inequality before. But not all that you have learned before that could be the suitable weapons in any given mathematical problem.

This question actually asked us try to solve using the suggested different representations and it then boils down to our capability to know when to stop when we realized something is amiss with our current inappropriate method.

I will get back to discuss of this problem at the end of this analysis. But the fact is that using only trigonometric identities will lead us to a standstill in the progress.

Therefore the answer for this part of the question is the combination of the first two suggested methods.

Question 4.

If you chose the "strictly trigonometric identities method", would you expand $\sin 9x$ as $\sin 9x = 3\sin 3x - 4(\sin^3 3x)$?



Again, we have to try that out before concluding for anything.

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

$3\sin 3x - 4(\sin^3 3x)-(1+\cot^2 x)=5\sin 3x+9\tan^2 x-1$

$-2\sin 3x - 4\sin^3 3x=9\tan^2 x+\cot^2 x$

$-2\sin 3x(1 +2\sin^2 3x)=9\tan^2 x+\cot^2 x$

Okay, the terms $9\tan^2 x+\cot^2 x$ and $1 +2\sin^2 3x)$ is always greater than zero, but these don't bring us any meaningful idea to continue, not to mention to solve for the equation.

Therefore, one is advised not to expand $\sin 9x$ using the triple angle formula for sine function as it brought us nowhere closer to the solution.

Question 5.

If you chosen the "AM-GM Inequality + Trigonometric Identities method", which of the following terms do you think would require the AM-GM inequality to handle them?

$\sin 9x$
$5\sin 3x$


AM-GM inequality tells us for both real $x, \,y>0$, then $\dfrac{x+y}{2}\ge \sqrt{x(y)}$, with equality when $x=y$.

That implicitly means AM-GM inequality is best used if we have terms and their reciprocals when they are all positive.

Notice if we convert the original equation by using the identity $1+\cot^2 x=csc^2 x$, we have:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

$\sin 9x=5\sin 3x+9\tan^2 x-1+\csc^2 x$

$\sin 9x=5\sin 3x+9\tan^2 x+\cot^2 x$

[MATH]\color{black}\sin 9x=5\sin 3x+\color{yellow}\bbox[5px,green]{9\tan^2 x+\dfrac{1}{\tan^2 x}}[/MATH]

Therefore, we can use AM-GM on [MATH]\color{yellow}\bbox[5px,green]{9\tan^2 x+\dfrac{1}{\tan^2 x}}[/MATH] as it will give us a rather useful result:

$\dfrac{9\tan^2 x+\dfrac{1}{\tan^2 x}}{2}\ge\sqrt{9\tan^2 x\left(\dfrac{1}{\tan^2 x}\right)}$

$\dfrac{9\tan^2 x+\dfrac{1}{\tan^2 x}}{2}\ge\sqrt{9\cancel{\tan^2 x}\left(\dfrac{1}{\cancel{\tan^2 x}}\right)}$

$\dfrac{9\tan^2 x+\dfrac{1}{\tan^2 x}}{2}\ge\sqrt{9}$

$9\tan^2 x+\dfrac{1}{\tan^2 x}\ge 2(3)$

$9\tan^2 x+\dfrac{1}{\tan^2 x}\ge 6$

with equality when $9\tan^2 x=\dfrac{1}{\tan^2 x}$, i.e. when $\tan x=\pm \dfrac{\pi}{6}$.

We have gone too far, but my point is to show you wholly how AM-GM could be useful in solving some of the equality.

And to answer for this question 5, the correct terms that required the AM-GM inequality to handle them are $9\tan^2x$ and $\csc^2x$.

Question 6.

What are the solutions for this problem?

$0,\,\dfrac{\pi}{3}(1+n)$, where $n$ is an integer.
$0,\,n\pi$, where $n$ is an integer.
$\dfrac{\pi}{6}(12-5n),\,\dfrac{\pi}{3}(1+n)$, where $n$ is an integer.
$\dfrac{\pi}{6}(12-5n),\,\dfrac{\pi}{6}(12-n)$, where $n$ is an integer.


I will show the full worked out solution to this problem here. My inclination is to use the "AM-GM Inequality+Trigonometric Identities" to tackle the given problem.

For the given equation which we then rewritten it such that:

$\sin 9x-\csc^2 x=5\sin 3x+9\tan^2 x-1$

$\sin 9x=5\sin 3x+9\tan^2 x-1+\csc^2 x$

$\sin 9x=5\sin 3x+9\tan^2 x+\cot^2 x$

$\sin 9x-5\sin 3x=9\tan^2 x+\cot^2 x$---(*)

In previous question, I have explained how AM-GM inequality tells us the sum of $9\tan^2 x+\dfrac{1}{\tan^2 x}=6$ when $9\tan^2 x=\dfrac{1}{\tan^2 x}$, i.e.  $x=\pi\left(2n-\dfrac{5}{6}\right),\,\pi\left(2n-\dfrac{1}{6}\right)$, where n is an integer.

But that is what our RHS of the equation (*) told us.

We have to study the LHS of the equation as well. Fortunately that is easy-peasy as we have:

$-1\le \sin 9x \le 1$

$-5\le -5\sin 3x \le 5$

$\therefore -6\le \sin 9x-5\sin 3x \le 6$

We have reached to a really exciting progress where the RHS of the equation says it is greater than or equal to 6 but the LHS of the equation says it must be less than or equal to 6.

The only sensible meaning that one could get from the above statement is the original equation holds only when both sides equal to 6.

Okay, when the LHS equals 6, we have:

1. $\tan x=\pm \dfrac{\pi}{6}$.

When the RHS equals 6, we see that

2. $\sin 9x=1$ and

3. $5\sin 3x=5\implies\,\,\sin 3x=1$

Condition in (1) is satisfied when $x=\pi\left(2n-\dfrac{5}{6}\right),\,\pi\left(2n-\dfrac{1}{6}\right)$, where n is an integer.

Condition in (2) is satisfied when $x=\dfrac{\pi}{18}(4n+\pi)$, where n is an integer.

Condition in (3) is satisfied when $x=\pi\left(\dfrac{2n}{3}-\dfrac{1}{6}\right),\,\pi\left(\dfrac{2n}{3}+\dfrac{1}{2}\right)$, where n is an integer.

Therefore, the solutions that exist for the given trigonometric equation are:

$x=\pi\left(2n-\dfrac{5}{6}\right),\,\pi\left(2n-\dfrac{1}{6}\right)$, where $n$ is an integer.

Question 7.

Please rate the difficulty on this problem on a scale from 1 to 5 with 5 being the highest level of difficulty and 1 being the lowest level of difficulty.


This is to check the overall capability from the students and to gather useful data so to crank out future quiz problems that are more appropriate to different abilities of students.

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