Triangle ABC: Find Possible Values of Cos A | POTW #449 Jan 4th 2021

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In summary, a triangle is a geometric shape with three sides and angles. The Cosine function, denoted as cos, is a math function that relates the side lengths of a right triangle to its angles. To find the possible values of Cos A for a triangle, you need to know the length of its sides and use the cosine function. The range of values for Cos A in a triangle is between -1 and 1, making it useful in solving geometry and trigonometry problems and determining properties of a triangle.
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anemone
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Here is this week's POTW:

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In a triangle $ABC$, let $\angle A\ge \angle B \ge \angle C$ and suppose that $\sin 4A+\sin 4B+\sin 4C=2(\sin 2A+\sin 2B+\sin 2C)$. Find all possible values of $\cos A$.

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As usual, I will give the community another week's time to attempt at last week's POTW. And I am looking forward to receiving submissions from the members!
 
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No one replied to last two week's POTW. (Sadface) However, you can read the official solution (by other) as below:
Using $A+B+C=\pi$, we have

$\begin{align*}\sin 2A+\sin 2B+\sin 2C&=\sin 2A+\sin 2B-\sin (2A+2B)\\&=2\sin (A+B)\cos (A-B)-2\sin(A+B)\cos(A+B)\\&=2\sin(A+B)[\cos(A-B)-\cos(A+B)]\\&=2\sin(A+B)(-2\sin A \sin B)\\&=2\sin C (2\sin A\sin B)\\&=4\sin A \sin B\sin C\end{align*}$

Replace $A,\,B$ and $C$ in the above calculation by $2A,\,2B$ and $2C$ to get

$\sin 4A+\sin 4B+\sin 4C=-4\sin 2A\sin 2B\sin 2C$

Using these two inequalities, the given relation is equivalent to

$-4cos A\cos B\cos C=1$

Since the product of cosines in $-4\cos A\cos B\cos C=1$ is negative, and $A\ge B\ge C$, we must have $A>\dfrac{\pi}{2}>2C>0$.

Using $A+B+C=\pi$, from $-4\cos A\cos B\cos C=1$, we get

$4\cos A \cos(A+C)\cos C=1$ and so

$4\cos A(\cos A \cos C-\sin A \sin C)\cos C=1$

Divide both sides by $\cos^2 C$ to get

$4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$

Let $x=\cos A$ and $t=\tan^2 C$.

From $A>\dfrac{\pi}{2}>2C>0$, it follows that $-1<x<0$ and $0<t<1$.

Rewrite $4\cos^2 A-4\sin A\cos A\tan C=\dfrac{1}{\cos^2 C}$ using this notation and rearrage to get

$4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$

Square both dies and move all the terms to one side to get

$16(1+t)x^4-8(1+3t)x^2+(t+1)^2=0$

Applying the quadratic formula, we get

$x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$

Since $-1<x<0$ and $0<t<1$, $4x^2-(1+t)<0$ whenever $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ holds.

For the solutions obtained from the quadratic formula we have

$4x^2-(1+t)=\dfrac{(t-t^2)\pm \sqrt{3t+6t^2-t^3}}{1+t}$ and since $t-t^2>0$ it is clear that when $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$, $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ is not satisfied.

Therefore, since $x<0$, we must have

$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$

We want to find the range of values of $x$ for $t\in(0,\,1)$. Implicitly differentiate $x^2=\dfrac{1+3t\pm\sqrt{3t+6t^2-t^3}}{4(1+t)}$ to get

$x\dfrac{dx}{dt}=\dfrac{4\sqrt{3t+6t^2-t^3}+t^3+3t^2-9t-3}{16(1+t)^2\sqrt{3t+6t^2-t^3}}$

There is no $t\in(0,\,1)$ for which either $x=0$ or $16(1+t)^2\sqrt{3t+6t^2-t^3}=0$, so we conclude that the critical points of $x$ satisfy

$4\sqrt{3t+6t^2-t^3}=-t^3-3t^2+9t+3\\(3+9t-3t^2-t^3)^2-16(3t+6t^2-t^3)=0\\t^6+6t^5-9t^4-44t^3-33t^2+6t+9=0\\(t-3)(t+1)^3(t^2+6t-30$

The only critical point in the range $[0,\,1]$ is $t=2\sqrt{3}-2$. The corresponding value of $x$, obtained after a tedious but straightforward calculation is $\dfrac{1=\sqrt{3}}{2}$. From

$x=-\dfrac{1}{2}\sqrt{\dfrac{1+3t-\sqrt{3t+6t^2-t^3}}{1+t}}$,

we easily evaluate

$\displaystyle \lim_{t\rightarrow 0^+} x=-\dfrac{1}{2}$ and

$\displaystyle \lim_{t\rightarrow 1^-} x=-\dfrac{\sqrt{2-\sqrt{2}}}{2}$

allowing us to conclude that for $t\in(0,\,1)$, we have

$-\dfrac{1}{2}<x\le \dfrac{1-\sqrt{3}}{2}$

Finally, we check that for each $x$ in this interval there is a corresponding triangle whose angles $A,\,B$ and $C$ satisfy the given relation. Suppose $x_0$ is such that $\dfrac{1}{2}<x_0\le \dfrac{1-\sqrt{3}}{2}$. Let $A=\cos ^{-1}x_0$. Since $\cos^{-1}$ is a decreasing function we have $A<\cos^{-1}(-0.5)=\dfrac{2\pi}{3}$.

By the intermediate value theorem, since $x$ is continuous on $0,\,2\sqrt{3}-3]$, there exists a $t+0$ in this interval such that $x_0=x(t_0)$.

Let $C=\tan^{-1}\sqrt{t_0}$. Note that $\sqrt{t_0}\le \sqrt{2\sqrt{3}-3}<\sqrt{3}$, where $C<\dfrac{\pi}{3}$.

Let $B=\pi-A-C$, the earlier comments about the ranges for $A$ and $C$ imply $B>0$.

We claim that a triangle with angles $A,\,B$ and $C$ satisfies the relation given in the problem. From the construction, $\cos C=(1+\tan^2 C)^{-0.5}=(1+t_0)^{-0.5}$ and $\cos A=x_0$. Moreover, $x_0$ and $t$ satisfy equation $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$. We calculate, using trigonometric equalities to evaluate $\sin A$ and $\sin C$.

$\begin{align*}\cos B&=\cos (\pi-(A+C))\\&=-\cos (A+C)\\&=-\cos A\cos C+\sin A \sin C\\&=-x_0\sqrt{\dfrac{1}{1+t_0}}+\sqrt{1-x_0^2}\cdot \sqrt{1-\dfrac{1}{1+t_0}}\\&=\sqrt{\dfrac{1}{1+t_0}}(-x_0+\sqrt{1-x_0^2}\cdot \sqrt{t_0})\end{align*}$

Since $x_0\ne 0$, we rearrange $4x^2-(1+t)=4x\sqrt{1-x^2}\sqrt{t}$ to get $\sqrt{1-x_0^2}\cdot \sqrt{t_0}=x_0-\dfrac{1+t_0}{4x_0}$.

Hence $\cos B=\sqrt{\dfrac{1}{1+t_0}}\cdot \dfrac{1+t_0}{-4x_0}=\dfrac{\cos C(1+t_0)}{-4\cos A}$.

It follows that $-4\cos A\cos B\cos C=1$, so $A,\,B,\,C$ satisfy $-4cos A\cos B\cos C=1$, which is equivalent to the equality in the question.

Therefore, we conclude that the possible range of value for $\cos A$ is given by $-\dfrac{1}{2}<\cos A\le \dfrac{1-\sqrt{3}}{2}$.
 

FAQ: Triangle ABC: Find Possible Values of Cos A | POTW #449 Jan 4th 2021

1. What is a triangle and how is it related to cosines?

A triangle is a three-sided polygon with three angles. The cosine function is a mathematical ratio that relates the lengths of the sides of a right triangle to its angles. In a right triangle, the cosine of an angle is equal to the adjacent side divided by the hypotenuse.

2. How do you find the possible values of cosine for angle A in triangle ABC?

To find the possible values of cosine for angle A, you need to know the lengths of the sides of triangle ABC. Then, you can use the cosine function (cos A = adjacent/hypotenuse) to calculate the value of cosine for angle A. The possible values of cosine will depend on the lengths of the sides of the triangle.

3. What is the range of possible values for cosine in a triangle?

The range of possible values for cosine in a triangle is between -1 and 1. Cosine is a trigonometric function and its values can never be less than -1 or greater than 1.

4. How do you determine the possible values of cosine if you only know the measures of the angles in a triangle?

If you only know the measures of the angles in a triangle, you can use the fact that the sum of the angles in a triangle is always 180 degrees. You can set up an equation using the cosine function and the given angle measures and solve for the possible values of cosine.

5. Can the value of cosine be negative in a triangle?

Yes, the value of cosine can be negative in a triangle. This can happen when the angle is in the second or third quadrant of the coordinate plane, where cosine is negative. It can also happen in a right triangle if the angle is obtuse, as the adjacent side will be negative in this case.

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