# Since 5i=5e^{i \frac{\pi}{2}} And 2+i=\sqrt5e^{ix_1} where x_1=\arctan \frac12, we have

Russell Gillen 2022-01-03
Since $$\displaystyle{5}{i}={5}{e}^{{{i}{\frac{{\pi}}{{{2}}}}}}$$ And $$\displaystyle{2}+{i}=\sqrt{{5}}{e}^{{{i}{x}_{{1}}}}$$ where $$\displaystyle{x}_{{1}}={\arctan{{\frac{{12}}{}}}}$$, we have $$\displaystyle{\frac{{{5}{i}}}{{{2}+{i}}}}=\sqrt{{5}}{\left[{\cos{{\left({\frac{{\pi}}{{{2}}}}-{x}_{{1}}\right)}}}+{i}{\sin{{\left({\frac{{\pi}}{{{2}}}}-{x}_{{1}}\right)}}}\right]}$$ Now from here how do I continue?

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Vasquez

$$LHS=\sqrt5 e^{i(\frac{\pi}{2}-x_1)},x_1=\arctan \frac12$$ (1)
Continue with
$$RHS=1+2i=\sqrt5 e^{ix_2}, \ x_2=\arctan 2$$ (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.

###### Not exactly what you’re looking for?
user_27qwe

Hint
$$\cos(\frac{\pi}{2}-x)=\sin x$$
$$\sin(\frac{\pi}{2}-x)=\cos x$$
$$\tan(\frac{\pi}{2}-x)=\frac{1}{\tan x}$$
So now let's develop the hint from where you left it i.e
$$\frac{5i}{1+2i}=\sqrt5(\cos(\frac{\pi}{2}-x)+i \sin(\frac{\pi}{2}-x))=a+ib$$
So $$a^2+b^2=5$$ and
$$\frac{b}{a}=\tan(\frac{\pi}{2}-x)=\frac{1}{\tan x}=2$$
And so $$5a^2=5$$ and $$a=\pm 1, \ b=\pm 2$$ 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),
$$\frac{5i}{2+i}=\frac{5e^{t \frac{\pi}{2}}}{\sqrt5 e^{t \arctan \frac12}}=\sqrt5 e^{i(\frac{\pi}{2}-\arctan \frac12)}=\sqrt5 \cos (\frac{\pi}{5}-\arctan \frac12)+i\sqrt5 \sin (\frac{\pi}{2}-\arctan \frac12)$$
Sketch the $$1,2,\sqrt5$$ right triangle and determine that the cosine of the relevant angle (which is complementary to the angle with tangent $$\frac12$$) is $$\frac{1}{\sqrt5}$$ and its sine is $$\frac{2}{\sqrt5}$$. The result above immediately simplifies to 1+2i.

nick1337

$$\begin{array}{}\cos(5x)\cos(3x)−\sin(3x)\sin(x)=\cos(2x) \\\frac12(\cos(2x)+\cos(8x))−\frac12(\cos(2x)−\cos(4x))=\cos(2x) \\\frac12(\cos(4x)+\cos(8x))−\cos(2x)=0 \\\cos(8x)+\cos(4x)−2\cos(2x)=0 \\2\cos(2x)\cos(6x)−2\cos(2x)=0 \\\cos(2x)[\cos(6x)−1]=0 \\\cos(2x)=0 \to x=\pm \frac{\pi}{4}+k\pi \end{array}$$
discarded because this solution will make zero the denominator of the original equation
$$\cos(6x)=1 \to x=\frac{k\pi}{3}$$