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Mathematics Question on argand plane

The amplitude of sin⁡π5+i(1−cos⁡π5)\sin \frac{\pi}{5} + i\left( 1 - \cos \frac{\pi}{5}\right)

A

2π5\frac{2\pi}{5}

B

π5\frac{\pi}{5}

C

π15\frac{\pi}{15}

D

π10\frac{\pi}{10}

Answer

π10\frac{\pi}{10}

Explanation

Solution

Let Z=sin⁡π5+i(1−cos⁡π5)Z = \sin \frac{\pi}{5} + i\left( 1 - \cos \frac{\pi}{5}\right)
Putting sin⁡π5=rcos⁡θ\sin \frac{\pi}{5} =r \cos \theta ...(i)
and 1−cos⁡π5=rsin⁡θ1 -\cos \frac{\pi}{5} = r \sin \theta...(ii)
∴    tan⁡θ=1−cos⁡π5sin⁡π5=2sin⁡2π102sin⁡π10cos⁡π10\therefore\:\:\:\: \tan \theta = \frac{1-\cos \frac{\pi}{5}}{\sin \frac{\pi}{5}} = \frac{ 2 \sin^{2} \frac{\pi}{10}}{2 \sin \frac{\pi}{10} \cos \frac{\pi}{10}}
⇒tan⁡θ=tan⁡π10⇒θ=π10\Rightarrow \tan \theta = \tan \frac{\pi}{10} \Rightarrow \theta = \frac{\pi}{10}