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I have a quick question about a difficult trigonometric functions problem that I have been assigned. The problem is as follows: Evaluate $$\cos36° - \cos72°$$ without the aid of a calculator. In terms of my attempts at the problem, I have converted $\cos 36°$ into 2$\cos^{2}18°$ - 1, and I have converted $\cos72°$ into 1 - 2$\sin^{2}36°$. By the property that $\sin x$ = $\cos(90° - x)$, the latter expression becomes 1 - 2$\cos^{2}54°$. Unfortunately, I've hit a roadblock and don't really know what to do from here. Would, perhaps, adding the two determined expressions yield anything of use? Thanks for all advice.

EDIT: Or, perhaps, would it be wiser to convert $\cos72°$ into 2$\cos^{2}36°$ - 1 and to convert $\cos36°$ into 1 - 2$\sin^{2}18°$?

Linear Pedant
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3 Answers3

8

Inscribe a regular pentagon inside the unit circle with one vertex at $(1,0)$, the other four vertices are located at $$(\cos 72^\circ, \pm \sin 72^\circ) \quad\text{ and }\quad (\cos 144^\circ, \pm \sin 144^\circ ) = (-\cos 36^\circ, \pm \sin 36^\circ)$$ By symmetry, the center of mass of the pentagon is located at the origin. Looking at the $x$-component alone, one find

$$1 + 2\cos72^\circ - 2\cos 36^\circ = 0 \quad\implies\quad \cos 36^\circ - \cos 72^\circ = \frac12$$

achille hui
  • 122,701
4

First write it as $$\cos 36°-\cos 72°=2\sin { \frac { 72°-36° }{ 2 } } \sin { \frac { 36°+72° }{ 2 } } =2\sin { 18°\sin { 54° } } =2\sin { 18° } \cos { 36° } $$ then let calculate $\sin { 18° } $ $$\cos { 36° } =\sin { 54° } \\ \cos { 2\cdot 18°= } \sin { 3\cdot 18° } \\ \theta =18°\\ \cos { 2\theta } =\sin { 3\theta } \\ 1-2\sin ^{ 2 }{ \theta } =3\sin { \theta -4\sin ^{ 3 }{ \theta } } \\ 4\sin ^{ 3 }{ \theta -2\sin ^{ 2 }{ \theta } -3\sin { \theta } +1 } =0\\ \left( \sin { \theta } -1 \right) \left( 4\sin ^{ 2 }{ \theta +2\sin { \theta } -1 } \right) =0$$ from here we get $\sin { 18°=\frac { \sqrt { 5 } -1 }{ 4 } } $

so

$$\cos 36°-\cos 72°=2\sin { \frac { 72°-36° }{ 2 } } \sin { \frac { 36°+72° }{ 2 } } =2\sin { 18°\sin { 54° } } =2\sin { 18° } \cos { 36° } =\\ =2\sin { 18°\left( 1-2\sin ^{ 2 }{ 18° } \right) } =2\sin { 18°-4\sin ^{ 3 }{ 18° } } =2\left( \frac { \sqrt { 5 } -1 }{ 4 } \right) -4{ \left( \frac { \sqrt { 5 } -1 }{ 4 } \right) }^{ 3 }\\ $$

haqnatural
  • 21,578
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You can compute $\cos72°$ with a bit of geometry or some algebra. Note that $\alpha=72°=360°/5$, so $x=\cos\alpha+i\sin\alpha$ is the fifth root of $1$ with the smaller argument. Since $x^4+x^3+x^2+x+1=0$, we also have $$ x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0 $$ that can also be rewritten as $$ \left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1=0 $$ that means $$ 4\cos\alpha^2+2\cos\alpha-1=0 $$ and so $$ \cos72°=\cos\alpha=\frac{-1+\sqrt{5}}{4} $$ Then $$ \cos36°=\sqrt{\frac{1+\cos72°}{2}}= \sqrt{\frac{2\sqrt{5}+6}{16}}=\frac{\sqrt{5}+1}{4} $$

egreg
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