# A tricky trigonometric problem

• sooyong94
In summary: Can both of them have magnitude 1?Oh, you are right. So it should be ##\sqrt{2}##. Thanks!In summary, a cubic equation is given and it is shown that x=1 is a root for all values of θ. It is also deduced that x-1 is a factor of the equation. By factoring the cubic and quadratic equations, the roots are found to be 1, cosθ, and sinθ, with the solutions being 1, 1/2, and √3/2 when θ=π/3. It is also found that for θ in the range of 0<=
sooyong94

## Homework Statement

A cubic equation is given as:
##x^{3} -(1+\cos \theta +\sin \theta)x^{2} +(\cos \theta \sin \theta +\cos \theta +\sin \theta)x-\sin \theta \cos \theta=0##

Show that x=1 is a root of the equation for all values of θ and deduce that x-1 is a factor to the above equation.

Hence, by factoring the cubic equation above, show that
##(x-1)[x^{2}-(\cos \theta +\sin \theta)x+\cos \theta \sin \theta]=0##

and the roots are given by
##1, \cos \theta, \sin \theta##
Write down the roots of the equation given that ##\theta=\frac{\pi}{3}##

Find all values of ##\theta## in the range ##0<=\theta<2\pi## such that two of the three roots are equal.

By considering ##\sin \theta -\cos \theta##, or otherwise, determine the greatest possible difference between the two roots, and find the values of ##\theta## for ##0<=\theta<2\pi## for which the two roots have the greatest difference.

## Homework Equations

Factor theorem, trigonometric equations

## The Attempt at a Solution

For the first part, I have plugged in x=1 and found that it is 0. Then I deduced that (x-1) is a factor.

I factored the cubic above and factored the quadratic, and the roots are 1, ##\cos \theta## and ##\sin \theta##. Then I plugged in ##\theta=\frac{\pi}{3}## and found out the solutions are
##1, \frac{1}{2}, \frac{sqrt{3}}{2}##. Is it correct?

Since the two roots are equal, therefore I set the following equations:
##\cos \theta =1##
##\sin \theta=1##
##\cos \theta=\sin \theta##

The values were found out to be 0, ##\frac{\pi}{4}##, ##\frac{\pi}{2}## and ##\frac{5\pi}{4}##. Hopefully I did not make any mistakes here... :P

For the last part, I need to consider ##\sin \theta-\cos \theta##. But how do I find the difference between the two roots?

Is it correct?
I cannot fault your method - did not check your arithmetic though.
I suspect that the only roots they want to be the same are the sin and cos ones - so you have overkill.

...how do I find the difference between the two roots?
If one root is a and the other is b, then the difference between the roots is |a-b|.
In your case the difference will depend on the angle.

Simon Bridge said:
I cannot fault your method - did not check your arithmetic though.
I suspect that the only roots they want to be the same are the sin and cos ones - so you have overkill.

If one root is a and the other is b, then the difference between the roots is |a-b|.
In your case the difference will depend on the angle.

Could it be 2?

Did you sketch out the graph of each root vs angle?

Yup. One is sqrt(2), and another two is 2.

Simon Bridge said:
I cannot fault your method - did not check your arithmetic though.
I suspect that the only roots they want to be the same are the sin and cos ones - so you have overkill.

No. there are three roots, 1, sinθ, cosθ, any pair of them can be equal.

Find all values of θ in the range 0<=θ<2π such that two of the three roots are equal.

ehild

sooyong94 said:
Yup. One is sqrt(2), and another two is 2.

I think, here you really need only the roots cosθ and sinθ. You are right, the greatest difference between them is √ 2.

ehild

ehild said:
I think, here you really need only the roots cosθ and sinθ. You are right, the greatest difference between them is √ 2.

ehild

Wasn't it should be 2? Since ##\sin \theta =1## and ##\cos \theta=1##?

For what values of ##\theta## is ##|\sin\theta-\cos\theta | = 2##?

sooyong94 said:
Wasn't it should be 2? Since ##\sin \theta =1## and ##\cos \theta=1##?

sin2θ+cos2θ=1. Can both of them have magnitude 1?

ehild

## 1. What is a tricky trigonometric problem?

A tricky trigonometric problem is a mathematical question that involves the use of trigonometric functions such as sine, cosine, and tangent to solve for unknown angles or sides in a triangle. These problems can be challenging and require a good understanding of trigonometry concepts and formulas.

## 2. How do I approach a tricky trigonometric problem?

There are several steps you can follow to approach a tricky trigonometric problem:1. Identify what is given and what is asked in the problem.2. Draw a diagram and label all known values and angles.3. Determine which trigonometric function(s) can be used to solve the problem.4. Set up the trigonometric equation(s) and solve for the unknown variable(s).5. Check your answer and make sure it makes sense in the context of the problem.

## 3. What are the common types of tricky trigonometric problems?

Some common types of tricky trigonometric problems include:1. Solving for unknown angles or sides in a right triangle.2. Using trigonometric identities to simplify complex expressions.3. Solving for the missing side or angle in an oblique triangle.4. Applying trigonometric functions in real-world problems, such as finding the height of a building or the angle of elevation.

## 4. What are some tips for solving tricky trigonometric problems?

Here are a few tips to help you solve tricky trigonometric problems:1. Familiarize yourself with basic trigonometric concepts and formulas.2. Practice drawing diagrams to visualize the problem.3. Break down the problem into smaller, manageable steps.4. Double-check your calculations and units as trigonometric problems often involve multiple steps.5. Practice, practice, practice!

## 5. How can I improve my trigonometry skills?

To improve your trigonometry skills, you can:1. Review and understand basic trigonometric concepts and formulas.2. Practice solving different types of trigonometric problems.3. Use online resources or textbooks for additional practice and examples.4. Seek help from a teacher or tutor if you are struggling with a specific concept.5. Apply trigonometry in real-world situations to see its practical applications.

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