What is the solution to this trigonometric challenge?

In summary, the Trigonometric Challenge is a mathematical puzzle or game that aims to test and improve one's understanding and application of trigonometric concepts. Anyone with a basic knowledge of trigonometry can participate, and it is commonly used in educational settings. The challenge typically involves a series of questions or problems that require the use of trigonometric formulas and techniques to solve. Participating in the challenge can improve problem-solving skills, analytical thinking, and understanding of trigonometric concepts. Resources such as textbooks, online tutorials, and practice problems are available to help prepare for the challenge, and seeking help from a teacher or tutor can also be beneficial.
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Evaluate $\dfrac{\sin^2 \dfrac{\pi}{7}}{\sin^4 \dfrac{2\pi}{7}}+\dfrac{\sin^2 \dfrac{2\pi}{7}}{\sin^4 \dfrac{3\pi}{7}}+\dfrac{\sin^2 \dfrac{3\pi}{7}}{\sin^4 \dfrac{\pi}{7}}$ without the help of a calculator.
 
Mathematics news on Phys.org
  • #2
Let $s_k=\sin \dfrac{k\pi}{7}$ and $c_k=\cos \dfrac{k\pi}{7}$.

We have

$\dfrac{\sin^2 x}{\sin^4 x}=\dfrac{1}{16\sin^2 x \cos^4 x}=\dfrac{1}{16\sin^2 x \cos^2 x}+\dfrac{1}{16\cos^4 x}=\dfrac{1}{16\sin^2 x}+\dfrac{1}{16\cos^2 x}+\dfrac{1}{16\cos^4 x}$

Therefore we want to find $\displaystyle \sum_{k=1}^3 \dfrac{s_k^2}{s_{2k}^4}=\sum_{k=1}^3 \dfrac{1}{16s_k^2}+\sum_{k=1}^3 \dfrac{1}{16c_k^2}+\sum_{k=1}^3 \dfrac{1}{16c_k^4}$

$\sin 7x=\sin x(64\sin^6 x-112\sin^4 x+56\sin^2 x-7)$

Let the polynomial $64x^6-112x^4+56x^2-7$ has roots $\pm s_1,\,\pm s_2,\,\pm s_3$ so that the polynomial $P_1(x)=64x^3-112x^2+56x-7$ has roots $s_1^2,\,s_2^2,\,s_3^2$ and the polynomial $Q_1(x)=7x^3-56x^2+112x-64$ has roots $s_1^{-2},\,s_2^{-2},\,s_3^{-2}$.

$\displaystyle \sum_{k=1}^3 \dfrac{1}{s_k^2}=s_1^{-2}+s_2^{-2}+s_3^{-2}=\dfrac{56}{7}=8$

And

$P_2(x)=-P_1(1-x)=64x^3-80x^2+24x-1$ has roots $c_1^2,\,c_2^2,\,c_3^2$ and the polynomial $Q_2(x)=x^3-24x^2+80x-64$ has roots $c_1^{-2},\,c_2^{-2},\,c_3^{-2}$.

$\displaystyle \sum_{k=1}^3 \dfrac{1}{c_k^2}=c_1^{-2}+c_2^{-2}+c_3^{-2}=24$

$\displaystyle \sum_{k=1}^3 \dfrac{1}{c_k^4}=c_1^{-4}+c_2^{-4}+c_3^{-4}=24^2-2\times 80=416$

$\therefore \displaystyle \sum_{k=1}^3 \dfrac{s_k^2}{s_{2k}^4}=\dfrac{8}{16}+ \dfrac{24}{16}+\dfrac{416}{16}=28.$
 

FAQ: What is the solution to this trigonometric challenge?

1. What is a trigonometric challenge?

A trigonometric challenge is a mathematical problem that involves the use of trigonometric functions, such as sine, cosine, and tangent, to solve for unknown angles or sides in a triangle.

2. What are the common types of trigonometric challenges?

The most common types of trigonometric challenges are solving for missing angles or sides in a right triangle, finding the area or perimeter of a triangle, and solving real-world problems involving triangles and trigonometric functions.

3. How do I solve a trigonometric challenge?

To solve a trigonometric challenge, you will need to use the appropriate trigonometric function, such as sine, cosine, or tangent, and apply the correct formula or identity to find the missing angle or side. It is also important to draw a diagram and label all known values to help visualize the problem.

4. What are some tips for solving trigonometric challenges?

Some tips for solving trigonometric challenges include memorizing the basic trigonometric identities, understanding the relationships between the sides and angles of a right triangle, and practicing with different types of problems to improve your skills.

5. Can technology be used to solve trigonometric challenges?

Yes, technology such as calculators and trigonometric tables can be used to solve trigonometric challenges. However, it is important to understand the concepts and formulas behind the solutions rather than relying solely on technology.

Similar threads

Replies
1
Views
854
Replies
2
Views
768
Replies
2
Views
842
Replies
1
Views
881
Replies
1
Views
836
Replies
7
Views
1K
Replies
1
Views
973
Replies
1
Views
886
Back
Top