Larry Riddle, Agnes Scott College

## Trigonometric Values for Special Angles

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