Sunday, June 14, 2015

Effective Teaching of Math Using Hard Trigonometry Contest Problem

Many of the so-called practices/drills that are designed to achieve the learning for mastery are just near aimless activities, and genuine illumination of important mathematical ideas is rare, not to mention providing the opportunities of making students to think deeply, creatively and critically in generating solutions. There is a near obsession with calculators/ICT which preventing students from gaining a true mastery of basic skills. Overall, these practice problems are watered-down and less interesting math activities.

Students don't gain anything from going over familiar ground. In fact, studies show that regardless of intellectual and problem solving skills level, all students did benefit from exposure to more advanced content/problems. They are ready to learn more advanced math concepts and solve intriguing math problems, as long as they are presented in an engaging, developmentally appropriate way.

Researchers support problem-solving approach in the math classroom because it engages students in inquiry, connect between ideas, prompting them to build on and improve their current knowledge as they construct explanations that help them solve the task at hand.

In order to know what challenging questions to use to move the math ideas forward, it is critical that teachers need to continually search for effective problems, albeit intriguing math problems are hard to come by.

The problem that will be discussed in this blog post is a really insightful and rich math contest problem.

I will not lead you to the solution so soon, instead, I will show you how the vast majority of students will attempt in their trials, as those are the common ways (and in fact their common practice) of solving trigonometric equations. There are nothing wrong with these common problem solving ways, just that those ways DO NOT shed any valuable insights and therefore are bad practices.

Solve the trigonometric equation.

$\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left((\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

Students, when they saw the LHS of the equation share the same argument, the would jump to do the following:

[MATH]\color{yellow}\bbox[5px,purple]{\sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)}[/MATH]

$=\sqrt{\sqrt{2}^2+\sqrt{6}^2}\cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}+\tan^{-1}\dfrac{\sqrt{6}}{\sqrt{2}}\right)$

$=\sqrt{8}\cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}+\tan^{-1} \sqrt{6}\right)$

$=\sqrt{8}\cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}+\tan^{-1} \sqrt{3}\right)$

$=\sqrt{8}\cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}+\dfrac{\pi }{3}\right)$

$=\sqrt{8}\cos \left(\dfrac{x}{5}+\dfrac{\pi }{4}\right)$

$=\sqrt{8}\left(\cos \left(\dfrac{x}{5}\right)\cos\left(\dfrac{\pi }{4}\right)-\sin \left(\dfrac{x}{5}\right)\sin\left(\dfrac{\pi }{4}\right)\right)$

$=\sqrt{8}\left(\cos \left(\dfrac{x}{5}\right)\left(\dfrac{1 }{\sqrt{2}}\right)-\sin \left(\dfrac{x}{5}\right)\left(\dfrac{1 }{\sqrt{2}}\right)\right)$

$=2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)$

Up to this point, we see we could not do anything further since that is the most simplified form for the LHS of the equation.

Now, for the RHS of the equation...[MATH]\color{yellow}\bbox[5px,green]{2\left(\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)}[/MATH]:

It is easy to see how our students would apply the angle sum and difference identities to get:

[MATH]\color{yellow}\bbox[5px,green]{2\left(\sin \left(\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left(\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)}[/MATH]

$\small=2\left(\left(\sin \left(\dfrac{x}{5}\right)\cos \left(\dfrac{2\pi}{3}\right)-\cos \left(\dfrac{x}{5}\right)\sin \left(\dfrac{2\pi}{3}\right)\right)-\,\,\,\,\,\,\,\,\,\,\,\,\left(\sin \left(\dfrac{3x}{5}\right)\cos \left(\dfrac{\pi}{6}\right)+\cos \left(\dfrac{3x}{5}\right)\sin \left(\dfrac{\pi}{6}\right)\right)\right)$

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

$=-\sin \left(\dfrac{x}{5}\right)-\sqrt{3}\cos \left(\dfrac{x}{5}\right)-\sqrt{3}\sin \left(\dfrac{3x}{5}\right)-\cos \left(\dfrac{3x}{5}\right)$

Students will then be inclined to use the triple angle formulas for both cosine and sine function so they get:

$=-\sin \left(\dfrac{x}{5}\right)-\sqrt{3}\cos \left(\dfrac{x}{5}\right)-\sqrt{3}\left(3\sin \left(\dfrac{x}{5}\right)-4\sin^3 \left(\dfrac{x}{5}\right)\right)-\,\,\,\,\,\,\,\,\,\,\,\,\left(4\cos^3 \left(\dfrac{x}{5}\right)-3\cos \left(\dfrac{x}{5}\right)\right)$

$=-(1+3\sqrt{3})\sin \left(\dfrac{x}{5}\right)-(\sqrt{3}-3)\cos \left(\dfrac{x}{5}\right)+4\sqrt{3}\sin^3 \left(\dfrac{x}{5}\right)-4\cos^3 \left(\dfrac{x}{5}\right)$

Putting what we have been simplified for both sides of the original trigonometric equation, we get:

$\small \sqrt{2} \cos \left(\dfrac{x}{5}-\dfrac{\pi }{12}\right)-\sqrt{6}\sin \left(\dfrac{x}{5}-\dfrac{\pi}{12}\right)=2\left(\sin \left((\dfrac{x}{5}-\dfrac{2\pi}{3}\right)-\sin \left((\dfrac{3x}{5}+\dfrac{\pi}{6}\right)\right)$

$\scriptsize 2\cos \left(\dfrac{x}{5}\right)-2\sin \left(\dfrac{x}{5}\right)=-(1+3\sqrt{3})\sin \left(\dfrac{x}{5}\right)-(\sqrt{3}-3)\cos \left(\dfrac{x}{5}\right)+4\sqrt{3}\sin^3 \left(\dfrac{x}{5}\right)-4\cos^3 \left(\dfrac{x}{5}\right)$

$\small (\sqrt{3}-3+2)\cos \left(\dfrac{x}{5}\right)+4\cos^3 \left(\dfrac{x}{5}\right)=(2-(1+3\sqrt{3}))\sin \left(\dfrac{x}{5}\right)+4\sqrt{3}\sin^3 \left(\dfrac{x}{5}\right)$

$(\sqrt{3}-1)\cos \left(\dfrac{x}{5}\right)+4\cos^3 \left(\dfrac{x}{5}\right)=(1-3\sqrt{3})\sin \left(\dfrac{x}{5}\right)+4\sqrt{3}\sin^3 \left(\dfrac{x}{5}\right)$

Up to this point, it does not seem like there exists an easy way out to solve for the angle $x$ for the equation that we have gotten above. It even looks like more complicated than the original given equation.

But we should not give up so easily and thought what we have tried carried little or no value.

We have to look back with clear mind and strike the unproductive method, and try another approach by using other trigonometry formulas. If we have encountered with intriguing hard math problem found it extremely challenging, perseverance will yield rich rewards, i.e. the more robust problem solving skills and the sense of satisfaction with our abilities.

If angle sum and difference identities are of no use, then we might want to try the sum-to-product formula for sine function that says $\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}$.

I will continue to finish the unfinished business in another blog post, last but not least, you're encouraged to solve this problem your own. And tell me if you could solve it with your way, okay?

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