what will be the value of Sin 2 (6)-sin 2 (12)+sin 2 (18)... till 15th term , angles are in degree.

what will be the value of Sin2(6)-sin2(12)+sin2(18)... till 15th term , angles are in degree.

$Sin^2(6)-sin^2(12)+sin^2(18)+.......+sin^2(6*15)\\ =Sin^2(6)-sin^2(12)+sin^2(18)+... +sin^2(42)-sin^2(90-42)....+sin^2(90-12)-sin^2(90-6)+sin^2(90)\\$

Before I go any further can you please assure me that you middle term is meant to have a minus sign in front of it. :)

what will be the value of Sin2(6)-sin2(12)+sin2(18)... till 15th term , angles are in degree.

$\small{ \begin{array}{|lcll|} \hline && \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\sin^2(48^{\circ})+\sin^2(54^{\circ})-\sin^2(60^{\circ})+\sin^2(66^{\circ})-\sin^2(72^{\circ})+\sin^2(78^{\circ})-\sin^2(84^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\sin^2(90^{\circ}-42^{\circ})+\sin^2(90^{\circ}-36^{\circ})-\sin^2(90^{\circ}-30^{\circ}) +\sin^2(90^{\circ}-24^{\circ}) \\ && -\sin^2(90^{\circ}-18^{\circ})+\sin^2(90^{\circ}-12^{\circ})-\sin^2(90^{\circ}-6^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\sin^2(12^{\circ})+\sin^2(18^{\circ})-\sin^2(24^{\circ})+\sin^2(30^{\circ})-\sin^2(36^{\circ})+\sin^2(42^{\circ}) \\ && -\cos^2(42^{\circ})+\cos^2(36^{\circ})-\cos^2(30^{\circ}) +\cos^2(24^{\circ})-\cos^2(18^{\circ})+\cos^2(12^{\circ})-\cos^2(6^{\circ})+\sin^2(90^{\circ}) \\ &=& \sin^2(6^{\circ})-\cos^2(6^{\circ})-\sin^2(12^{\circ})+\cos^2(12^{\circ})+\sin^2(18^{\circ})-\cos^2(18^{\circ})-\sin^2(24^{\circ})+\cos^2(24^{\circ}) \\ && +\sin^2(30^{\circ})-\cos^2(30^{\circ})-\sin^2(36^{\circ})+\cos^2(36^{\circ})+\sin^2(42^{\circ})-\cos^2(42^{\circ})+\sin^2(90^{\circ}) \\\\ && \boxed{~ \begin{array}{rcl} \cos(2\varphi) &=& \cos^2(\varphi)-\sin^2{\varphi}\\ -\cos(2\varphi) &=& \sin^2(\varphi)-\cos^2{\varphi}\\ \end{array} }\\ &=& -\cos(12^{\circ})+\cos(24^{\circ})-\cos(36^{\circ})+\cos(48^{\circ}) \\ && -\cos(60^{\circ})+\cos(72^{\circ})-\cos(84^{\circ})+\sin^2(90^{\circ}) \quad | \quad \sin^2(90^{\circ}) = 1^2=1\\ &=& 1 -\cos(12^{\circ})+\cos(24^{\circ})-\cos(36^{\circ})+\cos(48^{\circ}) -\cos^2(60^{\circ})+\cos(72^{\circ})-\cos(84^{\circ})\\ &=& 1 -\cos(12^{\circ})+\cos(24^{\circ})-\cos(60^{\circ}-24^{\circ})+\cos(60^{\circ}-12^{\circ}) \\ && -\cos(60^{\circ})+\cos(60^{\circ}+12^{\circ})-\cos(60^{\circ}+24^{\circ})\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-\cos(60^{\circ}-24^{\circ})-\cos(60^{\circ}+24^{\circ}) \\ && +\cos(60^{\circ}-12^{\circ})+\cos(60^{\circ}+12^{\circ})\\\\ && \boxed{~ \begin{array}{rcl} \cos(\alpha+\beta)+\cos(\alpha-\beta) &=& \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \\ && + \cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) \\ &=& 2 \cos(\alpha)\cos(\beta)\\ \end{array} }\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-[\cos(60^{\circ}-24^{\circ})+\cos(60^{\circ}+24^{\circ})] \\ && +[\cos(60^{\circ}-12^{\circ})+\cos(60^{\circ}+12^{\circ})]\\ &=& 1-\cos(60^{\circ}) -\cos(12^{\circ})+ \cos(24^{\circ})-2\cdot\cos(60^{\circ})\cos(24^{\circ}) +2\cdot \cos(60^{\circ})\cos(12^{\circ})\\\\ && \boxed{~ \begin{array}{rcl} \cos(60^{\circ}) &=& \frac12 \\ \end{array} }\\ &=& 1-\frac12 -\cos(12^{\circ})+ \cos(24^{\circ})-2\cdot\frac12\cos(24^{\circ}) +2\cdot \frac12\cos(12^{\circ})\\ &=& \frac12 -\cos(12^{\circ})+ \cos(24^{\circ})-\cos(24^{\circ}) + \cos(12^{\circ})\\ &=& \frac12 -\cos(12^{\circ})+ \cos(12^{\circ})+ \cos(24^{\circ})-\cos(24^{\circ}) \\ &\mathbf{=}& \mathbf{ \dfrac12 } \\ \hline \end{array} }$

Thanks Heureka :))