Solve for x. x^24+1=0 /Prove the distance formula for polar

[Fullmetal_Hye]

 

[pka]

The answer is easy: the cosine is an even function.
\(\displaystyle \L
\cos ( - \theta ) = \cos (\theta )\quad \Rightarrow \quad \cos \left( {\theta _1 - \theta _2 } \right) = \cos \left( {\theta _2 - \theta _1 } \right)\)

 

[soroban]

Re: Solve for x. x^24+1=0 /Prove the distance formula for po

Hello, Fullmetal_Hye!

I assume you're familar with DeMoivre's Theorem . . .

Solve for \(\displaystyle x\) and graph: \(\displaystyle \,x^{24}\,+\,1 \:=\:0\)

We have: \(\displaystyle \,x^{24} \:=\:-1\;\;\Rightarrow\;\;x \:=\-1)^{\frac{1}{24}}\)

We want the twenty-four 24^th roots of -1.

The polar form of -\(\displaystyle 1\) is: \(\displaystyle \:\cos(\pi + 2\pi n)\,+\,i\sin(\pi + 2\pi n)\)

Hence: \(\displaystyle \L\,(-1)^{\frac{1}{24}}\;=\;\cos\left(\frac{\pi\,+\,2\pi n}{24}\right) + i\sin\left(\frac{\pi\,+\,2\pi n}{24}\right)\;\) for \(\displaystyle n\;=\;0,1,2,3,\,...\.,23\)

Therefore, the roots are:

\(\displaystyle \;\;\;\begin{Bmatrix}x_1\:=\:\cos\frac{\pi}{24}\,+\,i\sin\frac{\pi}{24} \\ x_2\:=\:\cos\frac{3\pi}{24}\,+\,i\sin\frac{3\pi}{24} \\ x_3\:=\:\cos\frac{5\pi}{24}\,+\,i\sin\frac{5\pi}{24} \\ \vdots \\ x_{24}\:=\:\cos\frac{47\pi}{24}\,+\,i\sin\frac{47\pi}{24}\end{Bmatrix}\)


The first root is at approximately \(\displaystyle (x,y)\,=\,(0.99,\,0.13)\)
\(\displaystyle \;\;\)and the roots are spaced 15° apart on a unit circle.

Prove the distance formula for polar coordinates:

\(\displaystyle \;\;\;d \;= \;\sqrt{(r_1)^2\,+\,(r_2)^2\,-\,2r_1r_2\cos(\theta_1 - \theta_2)}\)

The fastest way is with the Law of Cosines.

             *(r1,θ1)
              \
            *   \
                  \d
         r1*        \
                      \
          *             *(r2,θ2)
           θ1-θ2    *
         *      *r2
            *
        *

We have: \(\displaystyle \,d^2\;=\;(r_1)^2 + (r_2)^2\,-\,2r_1r_2\cos(\theta_1-\theta_2)\)

Therefore: \(\displaystyle \L\,d\;=\;\sqrt{(r_1)^2\,+\,(r_2)^2\,-\,r_1r_2\cos(\theta_1-\theta_2)}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If they expect a proof done "the long way",

\(\displaystyle \;\;\)begin with: \(\displaystyle \,d \;=\;\sqrt{(x_2\,-\,x_1)^2\,+\,(y_2\,-\,y_1)^2}\)

Make these substitutions: \(\displaystyle \,\begin{Bmatrix}x_1\:=\:r_1\cos\theta_1,\;y_1\:=\:r_1\sin\theta_1 \\ x_2\:=\:r_2\cos\theta_2,\; y_2\:=\:r_2\sin\theta_2\end{Bmatrix}\)

Then do a lot of algebra and trig . . .