| Trig Value | Derivation | Back Button |
|---|---|---|
| \[\sin {54^ \circ } = \frac{{1 + \sqrt 5 }}{4}\] |
Let A=18°. Then 3A = 90- 2A, so cos(3A) = sin(2A). \[\begin{array}{l} 4 \cos ^3(A) - 3\cos (A) = 2\sin (A)\cos (A) \\ \Rightarrow 4{\cos ^2}(A) - 3 = 2\sin (A) \\ \Rightarrow 4 - 4{\sin ^2}(A) - 3 = 2\sin (A) \\ \Rightarrow 4{\sin ^2}(A) + 2\sin (A) - 1 = 0 \\ \Rightarrow \sin (A) = \frac{{ - 2 + \sqrt {20} }}{8} = \frac{{ - 1 + \sqrt 5 }}{4} \\ \Rightarrow \sin {54^ \circ } = \sin (3A) = 3\sin (A) - 4\sin ^3(A) = \frac{{1 + \sqrt 5 }}{4} \\ \end{array}\] |
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| \[\cos {36^ \circ } = \frac{{1 + \sqrt 5 }}{4}\] | See result for sin 54° = cos 36°. | |
| \[\sin {36^ \circ } = \sqrt {\frac{{5 - \sqrt 5 }}{8}} \] | \[{\sin ^2}{36^ \circ } = 1 - {\cos ^2}{36^ \circ } = 1 - {\left( {\frac{{1 + \sqrt 5 }}{4}} \right)^2} = \frac{{5 - \sqrt 5 }}{8}\] | |
| \[\cos {72^ \circ } = \frac{{\sqrt 5 - 1}}{4}\] | \[\begin{align} \cos {72^ \circ } &= \cos (2 \cdot {36^ \circ }) = 2{\cos ^2}{36^ \circ } - 1 \\ &= 2{\left( {\frac{{1 + \sqrt 5 }}{4}} \right)^2} - 1 = \frac{{\sqrt 5 - 1}}{4} \\ \end{align}\] | |
| \[\sin {72^ \circ } =\sqrt {\frac{{5 + \sqrt 5 }}{8}} \] | \[{\sin ^2}{72^ \circ } = 1 - {\cos ^2}{72^ \circ } = 1 - {\left( {\frac{{\sqrt 5 - 1}}{4}} \right)^2} = \frac{{5 + \sqrt 5 }}{8}\] | |
| \[\cos {108^ \circ } = \frac{{1 - \sqrt 5 }}{4}\] | \[\cos {108^ \circ } = \cos ({180^ \circ } - {72^ \circ }) = - \cos {72^ \circ } = \frac{{1 - \sqrt 5 }}{4}\] | |
| \[\sin {108^ \circ } = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \] | \[\sin {108^ \circ } = \sin ({180^ \circ } - {72^ \circ }) = \sin {72^ \circ } = \sqrt {\frac{{5 + \sqrt 5 }}{8}}\] | |
| \[\sin{A}+\cos{A}= \] \[\sqrt{2}\,\sin(A+\tfrac{\pi}{4})\] | \[\sin{A}+\cos{A} = \sqrt{2}\left( \sin{A}\cos{\tfrac{\pi}{4}}+\cos{A}\sin{\tfrac{\pi}{4}}\right)\] \[=\sqrt{2}\,\sin{(A+\tfrac{\pi}{4})}\] | |
| \[\cos{A}-\sin{A} = \] \[\sqrt{2}\,\cos{(A+\tfrac{\pi}{4})}\] | \[\cos{A}-\sin{A} = \sqrt{2}\left( \cos{A}\cos{\tfrac{\pi}{4}} - \sin{A}\sin{\tfrac{\pi}{4}} \right)\] \[= \sqrt{2}\,\cos\left(A+\tfrac{\pi}{4}\right) \] | |
| \[\sin\left(A+\tfrac{\pi}{4}\right) =\] \[-\cos\left(A+\tfrac{3\pi}{4}\right)\] | \[ \sin\left(A+\tfrac{\pi}{4}\right) = \tfrac{1}{\sqrt{2}}(\sin A + \cos A) =\] \[ -\left(\cos A\cos\tfrac{3\pi}{4} - \sin A\sin\tfrac{3\pi}{4}\right) = -\cos\left(A+\tfrac{3\pi}{4}\right) \] | |
| \[\cos\left(A+\tfrac{\pi}{4}\right) =\] \[\sin\left(A+\tfrac{3\pi}{4}\right)\] | \[ \cos\left(A+\tfrac{\pi}{4}\right) = \tfrac{1}{\sqrt{2}}(\cos A - \sin A) =\] \[ \left(\cos A\sin\tfrac{3\pi}{4} + \sin A\cos\tfrac{3\pi}{4}\right) = \sin\left(A+\tfrac{3\pi}{4}\right) \] |