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Question Number 121256 by ZiYangLee last updated on 06/Nov/20

Prove {2 \sin}^{2} 3 θ - {2 \sin}^{2} θ = \cos 2 θ - \cos 6 θ .

By substitusing θ = \frac{π}{10} in the above identity,

prove that \sin \frac{3 π}{10} - \sin \frac{π}{10} = \frac{1}{2}

Answered by TANMAY PANACEA last updated on 06/Nov/20

2 {s i n}^{2} 3 θ - 2 {s i n}^{2} θ

1 - c o s 6 θ - 1 + c o s 2 θ

c o s 2 θ - c o s 6 θ

s o s i n 3 θ - s i n θ = \frac{c o s 2 θ - c o s 6 θ}{2 (s i n 3 θ + s i n θ})

s i n \frac{3 π}{10} - s i n \frac{π}{10}

\frac{c o s \frac{2 π}{10} - c o s \frac{6 π}{10}}{2 (s i n \frac{3 π}{10} + s i n \frac{π}{10})}

\frac{1}{2} \times \frac{2 s i n (\frac{4 π}{10}) s i n (\frac{2 π}{20})}{2 s i n (\frac{2 π}{10}) c o s (\frac{π}{10})} = \frac{1}{2} p r o v e d

n o t e

s i n \frac{4 π}{10} = c o s (\frac{π}{2} - \frac{4 π}{10}) = c o s (\frac{π}{10})

Commented by ZiYangLee last updated on 08/Nov/20

thank you sir

Commented by TANMAY PANACEA last updated on 08/Nov/20

m o s t w e l c o m e

Answered by Dwaipayan Shikari last updated on 06/Nov/20

s i n \frac{3 π}{10} - s i n \frac{π}{10}

= \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = \frac{1}{2}