# The value of cos(pi/5)

Posted by: Gary Ernest Davis on: December 3, 2010

@capedsam Tweeted: “Can you find an exact value for cos(pi/5)? “
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It is actually fairly easy to find an exact value for $\cos (\frac{\pi}{5})$ by following a path through the complex numbers.
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To do this we use Euler’s identity $e^{ix}=\cos (x) + \sin (x)$ for all real numbers $x$.
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Jim Tanton (@jamestanton) has a very nice introduction to Euler’s identity here.
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As shorthand let’s write $c=\cos (\frac{\pi}{5}) \textrm{ and } s=\sin (\frac{\pi}{5})$.
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From Euler’s identity we know that:
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$e^{i\frac{\pi}{5}}=\cos (\frac{\pi}{5}) + \sin (\frac{\pi}{5}) =(c+is)^5$………(1)
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We raise both sides of equation (1) to the $5^{th}$ power:
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$(e^{i\frac{\pi}{5}})^5=(c+is)^5$……………………….(2)
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The left hand side of equation (2) is $(e^{i\frac{\pi}{5}})^5=e^{i\pi} =-1$.
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The right hand side of equation (2) is, by the binomial theorem:
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$(c+is)^5 = c^5+5ic^4s-10c^3s^2-10ic^2s^3+5cs^4+is^5$Â  ……..(3)
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We separate the expression in (3) into real and imaginary parts, and equate the real part to -1, and the imaginary part to 0:
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$c^5-10c^3s^2+5cs^4=-1$……………………………..(4)
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$5c^4s-10c^2s^3+s^5=0$……………………………….(5)
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We can substitute $1-s^2 \textrm{ for } c^2$ in (5) to get:
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$5(1-s^2)^2s-10(1-s^2)s^3+s^5=0$………….(6)
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Expanding (6) and dividing through by $s$ we get:
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$16s^4-20s^2+5=0$………………………………………(7)
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This is a quadratic equation for $s^2$ with roots $\frac{5+\sqrt{5}}{8}\textrm{ and } \frac{5-\sqrt{5}}{8}$
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This gives $c^2=1-s^2 = 1-\frac{5\pm\sqrt{5}}{8} = \frac{3\pm\sqrt{5}}{8}$
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Only one of these two roots can be equal to $s^2=\sin ^2(\frac{\pi}{5})$.
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Which one?
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We know $\sin (\frac{\pi}{5}) < \sin (\frac{\pi}{4}) =\frac{\sqrt{2}}{2}$ so $s^2 < \frac{2}{4}=\frac{1}{2}$.
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Therefore, $s^2= \frac{5-\sqrt{5}}{8}$ so $c^2=\frac{3+\sqrt{5}}{8}$ which means, since $\cos (\frac{\pi}{5})>0$, that:
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$cos (\frac{\pi}{5})=\sqrt{\frac{3+\sqrt{5}}{8}}$.
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One can also get the quadratic equation for $s^2$ from the multiple angle formulas (if one can remember them!).

### 4 Responses to "The value of cos(pi/5)"

Note also that cos(pi/5) = (1+sqrt(5))/4, which is half the golden ratio. Can you find a geometric proof?

Nice question. Thanks.
Of course $\sqrt{\frac{3+\sqrt{5}}{8}}=\frac{1+\sqrt{5}}{4}$

1). Connect all of the vertices of a regular pentagon.
2). Use similar triangle ratios to show that the sides of the 36-72-72 isosceles triangle are related by the golden ratio.
3). Use law of cosines to prove that cos(36) = phi/2

Here’s another: write sin 3pi/5 = sin 2pi/5, expand into sines and cosines of pi/5, and divide out the sin pi/5 which should show up in every term. You’re left with a quadratic in cos pi/5.