Question

Find the value of
$\sin(− 19π /3 )$
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Answer 1

Answer:

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Step-by-step explanation:

$so \: here \\ sin( \frac{ - 19\pi}{3} ) \\ \\ = - sin( \frac{19\pi}{3} ) \\ \\ = - sin(19 \times 60) \\ \\ = - sin(1140) \\ = - sin .1140 \\ so \: sin \: 1140 \: can \: be \: written \: as \: sin \: (1080 + 60) \\ ie \: sin(6\pi + \frac{\pi}{3} ) \\ \\ so \: we \: know \: that \\ whenever \: there \: is \: \pi \: in \: any \: trignometric \: ratio \: there \: is \: no \: change \\ \\ hence \: - sin.1140 = - sin \: 60 \\ = - \frac{ \sqrt{3} }{2} \\ since \: sin \: 60 = \frac{ \sqrt{3} }{2} \\ and \: as \: 1140 \: lies \: in \: first \: quadrant \: the \: sign \: would \: be \: positive \: \\ hence \: then \\ \\ sin( - \frac{19\pi}{3} ) = - \frac{ \sqrt{3} }{2}$