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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Expert Answer

Vasquez
Answered 2022-01-08 Author has 9499 answers

We already have
\(LHS=\sqrt5 e^{i(\frac{\pi}{2}-x_1)},x_1=\arctan \frac12\) (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.

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user_27qwe
Answered 2022-01-08 Author has 10046 answers

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 \(0 < \frac{\pi}{2}-x < \frac{\pi}{2}\)

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nick1337
Answered 2022-01-08 Author has 10160 answers

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.

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nick1337
Answered 2022-01-08 Author has 10160 answers

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

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