Solving Trigonometry Questions Using Analytic and Algebraic Techniques

Trigonometry, a branch of mathematics that studies relationships involving lengths and angles of triangles, often presents challenges in solving complex equations. This article delves into two intriguing trigonometry problems, providing a step-by-step solution and detailed explanations. Whether you are a high school student, a college math enthusiast, or a professional in a related field, this guide will provide valuable insights into solving these types of questions.

Problem 1: Analyzing Trigonometric Equations

The first problem involves two equations:

Equations:

Equation 1: (sin x - 3a m sin^3 a) Equation 2: (cos x - 3a m cos^3 a)

Our goal is to find the value of (cos x) and the possible values of (m).

Solution:

Let's start by setting ( sin a s ) and ( cos a c ).

Step 1: Express (s^2) and (c^2)

(s^2 c^2 1)

Step 2: Substitute and Simplify

(s^2c^6 3s^4c^2 3s^2c^4 s^6 1) (s^6 c^6 3s^2c^2(s^2 c^2) 1) (s^6 c^6 3s^2c^2 1) (s^6 c^6 1 - 3s^2c^2)

Step 3: Use Moivre's Theorem

(sin 3a 3c^2s - s^3) (cos 3a c^3 - 3cs^2)

Step 4: Express (cos x) and Solve for (m)

(cos x cos x - 3a 3a)

(cos x (cos x - 3a)cos 3a - (sin x - 3a)sin 3a)

(cos x m c^3(c^3 - 3cs^2) - m s^3(3c^2s - s^3))

(cos x m(c^6 - 3c^4s^2 - 3s^4c^2 s^6))

(cos x m(s^6 c^6 - 3s^2c^2))

(cos x m(1 - 6s^2c^2))

From (1 - 6s^2c^2 frac{1}{m^2}), we have:

(m^2 - 6 frac{1}{m^2})

(m^4 - 6m^2 - 1 0)

(m^2 3 pm 2sqrt{2})

(m pm sqrt{3 2sqrt{2}} pm 1 sqrt{2})

(m -sqrt{3 - 2sqrt{2}} -1 - sqrt{2})

Problem 2: Inverse Trigonometric Functions

The second problem involves finding the angle (theta) given the sides of a right triangle and using inverse trigonometric functions to solve for the angle.

Given Data:

For problem 10, (sec theta -frac{16}{13}), and (tan theta sqrt{sec^2 theta - 1} sqrt{256/169 - 1} sqrt{87}/13 approx 0.71749) For problem 11, (sin theta frac{4}{5})

Solutions:

Problem 10：

(theta) is in the third quadrant, so (tan theta > 0).

(tan theta frac{sqrt{87}}{13})

Problem 11：

(theta) is in the second quadrant, so (sin theta) and (csc theta) are positive, but (cos theta), (tan theta), (sec theta), and (cot theta) are negative.

(sin theta frac{4}{5})

(cos theta sqrt{1 - sin^2 theta} -frac{3}{5})

(tan theta frac{sin theta}{cos theta} -frac{4}{3})

(csc theta frac{1}{sin theta} frac{5}{4})

(sec theta frac{1}{cos theta} -frac{5}{3})

(cot theta frac{1}{tan theta} -frac{3}{4})

Conclusion

In this article, we have discussed two intriguing trigonometry problems and their solutions. By understanding the fundamental principles of trigonometric equations and inverse trigonometric functions, we can effectively solve complex problems. These techniques are crucial for students, mathematicians, and professionals in various fields, including engineering and physics, where precise calculations are essential.