Given,
[tex]\sf \bf :\longmapsto \sf sin6^0 sin42^0 sin56^0 Sin10[/tex]
Now we know a identity matching the statement . to use that, multiply and divide by 4
[tex]\sf \bf :\longmapsto \dfrac{1}{4}{ 2\sin {6}^{ \circ} \sin {66}^{ \circ} \times 2\sin {44}^{ \circ} \sin {102}^{ \circ} }[/tex]
[tex]\sf \bf :\longmapsto k=\dfrac{1}{4} \times 2 \sin {6}^{ \circ} \sin {66}^{ \circ} \times 2 \sin {42}^{ \circ} \sin {102}^{ \circ}[/tex]
[tex]\sf \bf : \red\longmapsto k = \dfrac{1}{4} \times (\cos {60}^{ \circ} - \cos {72}^{ \circ} ) (\cos {60}^{ \circ} - \cos {144}^{ \circ}[/tex]
[tex]\sf \bf :\longmapsto k = \dfrac{1}{4} \times \bigg( \dfrac{1}{2} - \dfrac{ \sqrt{5} - 1}{4} \bigg) \bigg( \dfrac{1}{2} + \cos {36}^{ \circ} \bigg)[/tex]
[tex]\sf \bf :\longmapsto k = \dfrac{1}{4} \times \bigg( \dfrac{1}{2} - \dfrac{ \sqrt{5} - 1}{4} \bigg) \bigg( \dfrac{1}{2} + \dfrac{ \sqrt{5} + 1}{4} \bigg)[/tex]
[tex]\sf \bf :\longmapsto k = \dfrac{1}{4} \times \bigg( \dfrac{ 3 - \sqrt{5} }{4} \bigg) \bigg( \dfrac{ 3 + \sqrt{5} }{4} \bigg):⟼k=41×(43−5)(43+5)
[/tex]
[tex]\sf \bf :\longmapsto k = \dfrac{1}{4} \times \bigg( \dfrac{ {3}^{2} - {( \sqrt{5} )}^{2} }{4 \times 4} \bigg):⟼k=41×(4×432−(5)2)[/tex]
[tex]\sf \bf :\longmapsto k=\dfrac{9-5}{16\times 4}[/tex]
[tex]\boxed{\sf \bf:\longmapsto\pink{ k=\dfrac{1}{16}}}[/tex]
Now , this is k, so to find k/2:
[tex]\sf \ :\longmapsto \green{ \dfrac{k}{2} = \dfrac{1}{2} \left(\dfrac{1}{16}\right)}[/tex]
[tex]\boxed{\boxed{\sf \bf :\longmapsto \dfrac{k}{2} = \dfrac{1}{32}}}[/tex]
Identities used:
[tex]\sf 2 \: sin \: A \: sin \: B = cos({A-B}[/tex]
[tex]\sf cos(180^0-A) = -cosAcos[/tex]
[tex]\sf cos72^0/\&\cos144^0 [/tex]
found using standard identities cos720 & cos1440 is found using standar
