Question

Is it possible to trisect an angle of$\mathbf{3}\mathbf{t}$ degrees if cos $\mathbf{3}\mathbf{t}\mathbf{=}\frac{1}{3}$? What if cos$\mathbf{3}\mathbf{t}\mathbf{=}\frac{11}{16}$?

Short answer

No, it is not possible to trisect both the angles.

Step-by-step solution

Conceptual introduction

Drawing precise lines, line segments, shapes, circles, and other figures with a ruler, compass, or protractor is known as geometric construction.

Step 2:Determine trisection of angle  3t

Recalling the values as:

The first one is:${\text{cos3t=4cos}}^{\text{3}}\text{t-3cost}$.

The second one is the rational root test.

If the value${\text{f(x)=a}}_{\text{n}}{\text{x}}^{\text{n}}\text{+}\mathrm{...}{\text{+a}}_{\text{1}}{\text{x+a}}_{\text{0}}$has a rational root$\text{r/s}$in the reduced form.

Then, getting:${\text{r|a}}_{\text{0}}$and${\text{s|}}_{\text{n}}$.

The angle of measure$\text{t}$may be constructed by trisecting an angle of$\text{3t}$, which is equal to creating the number.

It follows that${\text{4cos}}^{\text{3}}\text{t-3cost=1/3}$if.

This suggests that$12{\cos }^{3}t-9\cos t-1=0$.

According to the Rational Root Test mentioned above, the only choices for$\text{r}$are$\text{1}$, and the only possibilities for$\text{s}$are$3,-3,4,-4,12$and$\text{-12}$, if$\frac{r}{s}$is a rational solution to the aforementioned equation.

Therefore, the above equation can only have the following roots: .

$\frac{\text{1}}{\text{3}}\text{,}-\frac{\text{1}}{\text{3}}\text{,}\frac{\text{1}}{\text{4}}\text{,}\frac{-\text{1}}{\text{4}}\text{,}\frac{\text{1}}{\text{12}}\text{,}-\frac{\text{1}}{\text{12}}$

We can verify that none of these are in fact roots.

The supplied angle cannot be trisected as a result.

Further determine trisection of angle  3t

The angle of measure$t$may be constructed by trisecting an angle of$3t$, which is equal to creating the number.

If$\cos 3t=\frac{11}{16}$, then$4{\cos }^{3}t-3\cos t$must also equal$\frac{\text{11}}{\text{16}}$. Therefore,$64{\cos }^{3}t-48\cos t=11$.

According to the Rational Root Test mentioned above, the only options for$\text{r}$are$\text{1,-1,11}$and$\text{-11}$, and the only possibilities for$s$are$\frac{\text{1}}{{\text{2}}^{\text{n}}}$or$\text{0}\le \text{n}\le \text{6}$if$\text{r/s}$is a rational solution to the aforementioned equation.

The aforementioned equation may be simplified as$64{\left(\frac{11}{2^n}\right)}^3=48\left(\frac{11}{2^n}\right)+11$.

It isif:$r=\pm 11$.

Add${\text{2}}^{\text{3n}}$to both sides. The result is$\text{64}{\left(\text{11}\right)}^{\text{3}}=\text{48}\left(\text{11}\right)\left({\text{2}}^{\text{2n}}\right)+\left(\text{11}\right)\left({\text{2}}^{\text{3n}}\right)$.

While the right side is not divisible by${\text{(11)}}^{\text{3}}$, the left side is.

The equation cannot be fulfilled as a result.

Thus, getting$\text{r}\ne \text{±11}$:.

The aforementioned equation becomes$\text{64}{\left(\frac{\text{11}}{{\text{2}}^{\text{n}}}\right)}^{\text{3}}=\text{48}\left(\frac{\text{11}}{{\text{2}}^{\text{n}}}\right)\text{+11}$.

It isif:$r=\pm 1$.

Add${\text{2}}^{\text{3n}}$to both sides. It then returns$64{\left(11\right)}^3=48\left(11\right)\left(2^{2n}\right)+\left(11\right)\left(2^{3n}\right)$.

Consider the left side is not a power of$\text{2}$, the right side is. The equation cannot be fulfilled as a result.

Thus,$\text{r}\ne \text{±1}$.

As a result, all potential values for $\text{r}$ are ruled out. The angle cannot be trisected as a result since the equation cannot hold in any circumstance.