Hardest Trigonometric Equation (2)

In today post, I will take up where we left off (effective-teaching-of-math), to continue solving the trigonometric equation

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

The LHS of the equation has been simplified down to

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.

Method 2: Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$. Find $x,\,y$.

Method 2:

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?

Yes.
No.
Perhaps.

Answer:

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.

One Worked Example of Solving Heuristic IMO Math Problem

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.

The solution from the U.K. mathematician:

Olympiad Algebra Problem: Evaluate $ax^5+by^5$

Olympiad Algebra Problem:

If $a,\,b,\,x,\,y\in R$ such that

$ax +by =7$

$ax^2+by^2=49$

$ax^3+by^3=133$

$ax^4+by^4=406$

Evaluate $ax^5+by^5$.