Question

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Answer 1

$hello \: friend \\ here \: is \: your \: answer\\ the \: question \: is \: to \: find \: the \\ value \: of \: \sin(2x) \\ the \: question \: is \: \\ cos2x = cos60.cos30 + sin60 \: sin30$

$we \: know \: that \: the \: relation \\ between \: \: \sin( \alpha ) \: \: and \: \cos( \alpha ) \\ \sin( \alpha ) = \cos(90 - \beta ) \\ so \: here \: \\ \sin(2x) = \sin(90 - 2x) \\ = \cos(2x) \\ now \\ \cos(60) \\ = \cos(90 - 60) \\ = \sin(30) \\ \cos(30) \\ = \cos(90 - 60) \\ = \sin(30) \\ so \: we \: can \: write \\ cos2x = \cos60 \cos30 + \sin60 \sin30 \\ = > \sin2x = sin30sin60 + sin60.sin30 \\ \\ we \: know \: that \: the \: value \: of \\ \sin(30) = \frac{1}{2} \\ \sin(60) = \frac{ \sqrt{3} }{2} \\ \sin \: 30sin60 + \sin60. \sin30 \\ = \frac{1}{2} \times \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{2} \times \frac{1}{2} \\ = \frac{ \sqrt{3} }{4} + \frac{ \sqrt{3} }{4} \\ = \frac{ 2\sqrt{3} }{4} \\ = \frac{ \sqrt{3} }{2} \\ so \: we \: get \: that \: the \: vaue \: of \\ \\ = > \sin(2x) = \frac{ \sqrt{3} }{2}$

✨ hope it helps✨✨