Russell Gillen

2022-01-03

Since $5i=5{e}^{i\frac{\pi }{2}}$ And $2+i=\sqrt{5}{e}^{i{x}_{1}}$ where ${x}_{1}=\mathrm{arctan}\frac{12}{}$, we have $\frac{5i}{2+i}=\sqrt{5}\left[\mathrm{cos}\left(\frac{\pi }{2}-{x}_{1}\right)+i\mathrm{sin}\left(\frac{\pi }{2}-{x}_{1}\right)\right]$ Now from here how do I continue?

Vasquez

$LHS=\sqrt{5}{e}^{i\left(\frac{\pi }{2}-{x}_{1}\right)},{x}_{1}=\mathrm{arctan}\frac{1}{2}$ (1)
Continue with
(2)
${x}_{1},{x}_{2}$ are complementary angles, ${x}_{2}=\frac{\pi }{2}-{x}_{1}\to LHS=RHS$
Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since r and $\theta$ are same for LHS and RHS.

user_27qwe

Hint
$\mathrm{cos}\left(\frac{\pi }{2}-x\right)=\mathrm{sin}x$
$\mathrm{sin}\left(\frac{\pi }{2}-x\right)=\mathrm{cos}x$
$\mathrm{tan}\left(\frac{\pi }{2}-x\right)=\frac{1}{\mathrm{tan}x}$
So now let's develop the hint from where you left it i.e
$\frac{5i}{1+2i}=\sqrt{5}\left(\mathrm{cos}\left(\frac{\pi }{2}-x\right)+i\mathrm{sin}\left(\frac{\pi }{2}-x\right)\right)=a+ib$
So ${a}^{2}+{b}^{2}=5$ and
$\frac{b}{a}=\mathrm{tan}\left(\frac{\pi }{2}-x\right)=\frac{1}{\mathrm{tan}x}=2$
And so $5{a}^{2}=5$ and The negative solutions are to be excluded because $0<\frac{\pi }{2}-x<\frac{\pi }{2}$

nick1337

Using polar form (with exponential notation for convenience),
$\frac{5i}{2+i}=\frac{5{e}^{t\frac{\pi }{2}}}{\sqrt{5}{e}^{t\mathrm{arctan}\frac{1}{2}}}=\sqrt{5}{e}^{i\left(\frac{\pi }{2}-\mathrm{arctan}\frac{1}{2}\right)}=\sqrt{5}\mathrm{cos}\left(\frac{\pi }{5}-\mathrm{arctan}\frac{1}{2}\right)+i\sqrt{5}\mathrm{sin}\left(\frac{\pi }{2}-\mathrm{arctan}\frac{1}{2}\right)$
Sketch the $1,2,\sqrt{5}$ right triangle and determine that the cosine of the relevant angle (which is complementary to the angle with tangent $\frac{1}{2}$) is $\frac{1}{\sqrt{5}}$ and its sine is $\frac{2}{\sqrt{5}}$. The result above immediately simplifies to 1+2i.

nick1337

$\begin{array}{}\mathrm{cos}\left(5x\right)\mathrm{cos}\left(3x\right)-\mathrm{sin}\left(3x\right)\mathrm{sin}\left(x\right)=\mathrm{cos}\left(2x\right)\\ \frac{1}{2}\left(\mathrm{cos}\left(2x\right)+\mathrm{cos}\left(8x\right)\right)-\frac{1}{2}\left(\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(4x\right)\right)=\mathrm{cos}\left(2x\right)\\ \frac{1}{2}\left(\mathrm{cos}\left(4x\right)+\mathrm{cos}\left(8x\right)\right)-\mathrm{cos}\left(2x\right)=0\\ \mathrm{cos}\left(8x\right)+\mathrm{cos}\left(4x\right)-2\mathrm{cos}\left(2x\right)=0\\ 2\mathrm{cos}\left(2x\right)\mathrm{cos}\left(6x\right)-2\mathrm{cos}\left(2x\right)=0\\ \mathrm{cos}\left(2x\right)\left[\mathrm{cos}\left(6x\right)-1\right]=0\\ \mathrm{cos}\left(2x\right)=0\to x=±\frac{\pi }{4}+k\pi \end{array}$
discarded because this solution will make zero the denominator of the original equation
$\mathrm{cos}\left(6x\right)=1\to x=\frac{k\pi }{3}$

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