A peculiar form

Algebra Level 3

cosh ( 2 ) + i sin ( x ) = 1 e 2 \large \cosh{(2)}+i\sin{\left(\sqrt{x}\right)}=\frac{1}{e^2} Find x x .


The answer is -4.

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2 solutions

Chew-Seong Cheong
Mar 18, 2017

cosh 2 + i sin x = 1 e 2 e 2 + e 2 2 + i i ( e x i e x i ) 2 = e 2 Multiplying both sides by 2 e 2 + e 2 e x i + e x i = 2 e 2 e x i e x i = e 2 e 2 x i = 2 Squaring both sides x = 4 Note that i 2 = 1 x = 4 \begin{aligned} \cosh 2 + i\sin \sqrt x & = \frac 1{e^2} \\ \frac {e^2+e^{-2}}2 + i \cdot \frac {i(e^{-\sqrt xi}-e^{\sqrt xi})}2 & = e^{-2} & \small \color{#3D99F6} \text{Multiplying both sides by }2 \\ e^2+e^{-2} - e^{-\sqrt xi}+e^{\sqrt xi} & = 2e^{-2} \\ e^{\sqrt xi} - e^{-\sqrt xi} & = e^{-2} - e^2 \\ \implies \sqrt xi & = - 2 & \small \color{#3D99F6} \text{Squaring both sides} \\ -x & = 4 & \small \color{#3D99F6} \text{Note that }i^2 = -1 \\ \implies x & = \boxed{-4} \end{aligned}

Nice solution!

Akeel Howell - 4 years, 2 months ago

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Thanks. I have shortened it further.

Chew-Seong Cheong - 4 years, 2 months ago
Akeel Howell
Mar 18, 2017

We have that cosh ( 2 ) + i sin ( x ) = 1 e 2 \displaystyle\cosh{(2)}+i\sin{\left(\sqrt{x}\right)}=\dfrac{1}{e^2} . If we rewrite 1 e 2 = e 2 \dfrac{1}{e^2} = e^{-2} as cos θ + i sin θ \cos{\theta}+i\sin{\theta} , we see that e 2 = e i ( 2 i ) e^{-2} = e^{i(2i)} . From this we have θ = 2 i \theta = 2i

cos ( 2 i ) = e i ( 2 i ) + e i ( 2 i ) 2 = e 2 + e 2 2 = cosh ( 2 ) cosh ( 2 ) + i sin ( 2 i ) = cos ( 2 i ) + i sin ( 2 i ) \cos{(2i)} = \dfrac{e^{i(2i)}+e^{-i(2i)}}{2} = \dfrac{e^{-2}+e^{2}}{2} =\ \cosh{(2)} \\ \therefore \cosh{(2)}+i\sin{(2i)} = \cos{(2i)}+i\sin{(2i)}

Hence, we see that x = 2 i x = 4 \sqrt{x} = 2i \implies x = \boxed{-4}

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