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\[\sin {54^ \circ } = \frac{{1 + \sqrt 5 }}{4}\]
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Let A=18°. Then 3A = 90- 2A, so cos(3A) = sin(2A).
Expanding each of these trig functions gives
\[\begin{array}{l}
4{\cos ^2}(A) - 3\cos (A) = 2\sin (A)\cos (A) \\
\Rightarrow 4{\cos ^2}(A) - 3 = 2\sin (A) \\
\Rightarrow 4 - 4{\sin ^2}(A) - 3 = 2\sin (A) \\
\Rightarrow 4{\sin ^2}(A) + 2\sin (A) - 1 = 0 \\
\Rightarrow \sin (A) = \frac{{ - 2 + \sqrt {20} }}{8} = \frac{{ - 1 + \sqrt 5 }}{4} \\
\Rightarrow \sin {54^ \circ } = \sin (3A) = 3\sin (A) - 4{\sin ^2}(A) = \frac{{1 + \sqrt 5 }}{4} \\
\end{array}\]
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\[\cos {36^ \circ } = \frac{{1 + \sqrt 5 }}{4}\]
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See result for sin 54° = cos 36°.
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\[\sin {36^ \circ } = \sqrt {\frac{{5 - \sqrt 5 }}{8}} \]
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\[{\sin ^2}{36^ \circ } = 1 - {\cos ^2}{36^ \circ } = 1 - {\left( {\frac{{1 + \sqrt 5 }}{4}} \right)^2} = \frac{{5 - \sqrt 5 }}{8}\]
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\[\cos {72^ \circ } = \frac{{\sqrt 5 - 1}}{4}\]
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\[\begin{align}
\cos {72^ \circ } &= \cos (2 \cdot {36^ \circ }) = 2{\cos ^2}{36^ \circ } - 1 \\
&= 2{\left( {\frac{{1 + \sqrt 5 }}{4}} \right)^2} - 1 = \frac{{\sqrt 5 - 1}}{4} \\
\end{align}\]
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\[\sin {72^ \circ } =\sqrt {\frac{{5 + \sqrt 5 }}{8}} \]
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\[{\sin ^2}{72^ \circ } = 1 - {\cos ^2}{72^ \circ } = 1 - {\left( {\frac{{\sqrt 5 - 1}}{4}} \right)^2} = \frac{{5 + \sqrt 5 }}{8}\]
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\[\cos {108^ \circ } = \frac{{1 - \sqrt 5 }}{4}\]
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\[\cos {108^ \circ } = \cos ({180^ \circ } - {72^ \circ }) = - \cos {72^ \circ } = \frac{{1 - \sqrt 5 }}{4}\]
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\[\sin {108^ \circ } = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \]
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\[\sin {108^ \circ } = \sin ({180^ \circ } - {72^ \circ }) = \sin {72^ \circ } = \sqrt {\frac{{5 + \sqrt 5 }}{8}}\] |
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