A tricky trigonometric problem

[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?

 

[Simon Bridge]

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.

 

[sooyong94]

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?

 

[Simon Bridge]

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

 

[sooyong94]

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

 

[ehild]

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

 

[ehild]

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

 

[sooyong94]

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 $\cos \theta=1$?

 

[Simon Bridge]

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

 

[ehild]

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