An algebra problem by Sandeep Bhardwaj

Algebra Level 3

( cos 9 x i sin 9 x ) 10 ( cos 10 x + i sin 10 x ) 11 ( cos 10 x + i sin 10 x ) 20 ( cos 18 x i sin 18 x ) 10 = ? \dfrac{ \left( \cos 9x -i \sin 9x \right)^{10} \left( \cos 10x + i \sin 10x \right)^{11} }{\left( \cos 10x +i\sin 10x \right)^{20} \left( \cos 18x -i \sin 18x \right)^{10}}= \, ?

Clarification : i = 1 i=\sqrt{-1} .


The answer is 1.

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1 solution

Rishabh Jain
Mar 15, 2016

Expression can be re written as:- ( cos ( 9 x ) + i sin ( 9 x ) ) 10 ( cos 10 x + i sin 10 x ) 11 ( cos 10 x + i sin 10 x ) 20 ( cos ( 18 x ) + i sin ( 18 x ) ) 10 \dfrac{ \left( \cos (-9x) +i \sin (-9x) \right)^{10} \left( \cos 10x + i \sin 10x \right)^{11} }{\left( \cos 10x +i\sin 10x \right)^{20} \left( \cos (-18x) +i \sin (-18x) \right)^{10}} Using Euler's Formula , expression can be simplified as:- = ( e 9 x i ) 10 ( e 10 x i ) 11 ( e 10 x i ) 20 ( e 18 x i ) 10 = ( e 90 x i ) ( e 110 x i ) ( e 200 x i ) ( e 180 x i ) = e 20 x i e 20 x i = 1 \large \begin{aligned} & =\dfrac{\left(e^{-9xi}\right)^{10}\left(e^{10xi}\right)^{11}}{\left(e^{10xi}\right)^{20}\left(e^{-18xi}\right)^{10}}\\&\\&=\dfrac{\left(e^{-90xi}\right)\left(e^{110xi}\right)}{\left(e^{200xi}\right)\left(e^{-180xi}\right)}\\&\\&=\dfrac{\cancel{e^{20xi}}}{\cancel{e^{20xi}}}=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{1}}}}}\end{aligned}

Cool, thanks!

Sandeep Bhardwaj - 5 years, 2 months ago

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Quite late....... BTW Welcome.... ;-)

Rishabh Jain - 5 years, 2 months ago

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