Question

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Answer 1

$\displaystyle\bf\\Sin\alpha-Sin2\alpha +Sin 3\alpha =(Sin\alpha+Sin3\alpha )-Sin 2\alpha =\\\\\\=2Sin\frac{\alpha +3\alpha }{2} Cos\frac{\alpha -3\alpha }{2} -Sin2\alpha =2Sin2\alpha Cos2\alpha -Sin2\alpha =\\\\\\=Sin4\alpha -Sin2\alpha \\\\1) \ \ \alpha =15^\circ\\\\Sin(4\cdot 15^\circ)-Sin(2\cdot 15^\circ)=Sin60^\circ-Sin30^\circ=\\\\\\=\frac{\sqrt{3} }{2} -\frac{1}{2} =\frac{\sqrt{3}-1 }{2}$

$\displaystyle\bf\\2) \ \ \alpha =30^\circ\\\\Sin(4\cdot 30^\circ)-Sin(2\cdot 30^\circ)=Sin120^\circ-Sin60^\circ=\\\\\\=\frac{\sqrt{3} }{2} -\frac{\sqrt{3} }{2} =0\\\\3) \ \ 2) \ \ \alpha =90^\circ\\\\Sin(4\cdot 90^\circ)-Sin(2\cdot 90^\circ)=Sin360^\circ-Sin180^\circ=\\\\\\=0-0=0$