Question

Prove that 4sin27 = (5+√5)¹/² - (3- √5)¹/²

Answer 1

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$we \: know \: that \: \\ \\ = > sin(54) = cos(90 - 54) \: \: \: \: \: \: \: ...(since \: = > cos(90 - \alpha ) = sin \alpha ) \\ \\ = > sin(54) = cos(36) \\ \\ = > sin54 = \frac{1}{4} ( \sqrt{5} + 1) \\ \\ - - - - - - - - - - - - - \\ we \: know \: that \\ = > {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ = > {cos}^{2} \alpha = 1 - {sin}^{2} \alpha \\ = > cos \alpha = \sqrt{1 - {sin}^{2} \alpha } \\ - - - - - - - - - - - - - \\ \\ here \: = > \alpha = 54 \\ \\ plug \: the \: value \: of \: = > \alpha = 54 \\ \\ we \: get \\ - - - - - - - - - - - - - - \\ \\ = > cos54 = \sqrt{1 - {sin}^{2}54 } \\ \\ = > cos54 = \sqrt{1 - {( \frac{1}{4} ( \sqrt{5 } +1)) }^{2} } \\ \\ = > cos54 = \sqrt{1 - \frac{1}{16} (5 + 1 + 2 \sqrt{5} )} \\ \\ = > cos54 = \frac{1}{4} ( \sqrt{16 - 5 - 1 - 2 \sqrt{5} } \: ) \\ \\ = > cos54 = \frac{1}{4} \sqrt{10 - 2 \sqrt{5} } \: \: \: ......eq _{1} \\ - - - - - - - - - - - - - - \\ \\ = > cos54 = cos2(27) \\ \\ = > cos54 = 1 - 2 {sin}^{2} 27 \: \: \: \: \: \: \: \: \: (cos2 \alpha \: = 1 - 2 {sin}^{2} \alpha ) \\ \\ - - - - - - - - - - - - - - - - \\ \\ from \: eq _{1} \\ \\ = > 1 - 2 {sin}^{2} 27 = \frac{1}{4} \sqrt{10 - 2 \sqrt{5} } \\ \\ = > 2 {sin}^{2} 27 = 1 - \frac{1}{4} \sqrt{10 - 2 \sqrt{5} } \\ \\ multiply \: both \: side \: by \: 8 \: we \: get \\ \\ = > 16 {sin}^{2} 27 = 8(1 - \frac{1}{4} \sqrt{10 - 2 \sqrt{5} } ) \\ \\ = > {(4sin27) }^{2} = 8 - 2 \sqrt{10 - 2 \sqrt{5} } \\ \\ = > {(4sin27)}^{2} = { (\sqrt{(5 + \sqrt{5} } ) - \sqrt{(3 - \sqrt{5} }) })^{2} \\ \\ = > 4sin27 \: = \: \sqrt{5 + \sqrt{5} } - \sqrt{3 - \sqrt{5} } \: \: \: \: \: \: \: ....proved$


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