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The amplitude of $\; sin \large\frac{\pi}{5} + i\; \normalsize cos (1-\large\frac{\pi}{5})\;$ is


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Answer : $\;\large\frac{\pi}{10}$
Explanation :
We have , $\; sin \large\frac{\pi}{5} + i\; \normalsize cos (1-\large\frac{\pi}{5})\;$
$r cos \theta = sin \large\frac{\pi}{5}\;$ and $\; r sin \theta = 1- cos \large\frac{\pi}{5}$
$tan \theta = \large\frac{1 - cos \large\frac{\pi}{5}}{sin \large\frac{\pi}{5}}$
$= \large\frac{2 sin^{2} (\large\frac{\pi}{10})}{2 sin (\large\frac{\pi}{10}) (cos \large\frac{\pi}{10})}$
$tan \theta = \large\frac{\pi}{10} \qquad [\theta = \large\frac{\pi}{10}]$
answered Apr 19, 2014 by yamini.v

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