# Easier way to show $\cos(20^\circ)\cos(30^\circ)\cos(40^\circ)=\cos^2(10^\circ)\cos(50^\circ)$

The only way I can show $$\cos(20^\circ)\cos(30^\circ)\cos(40^\circ)=\cos^2(10^\circ)\cos(50^\circ)$$ stinks of sweat and brute force:

If $\cos(10^\circ)=x$ then using $\cos(a+b) = \cos a \cos b - \sin a \sin b$ and $\cos^2 + \sin^2 = 1$ you get $$\cos(20^\circ) = 2x^2-1\\ \cos(30^\circ) = 4x^3-3x \\ \cos(40^\circ) = 8x^4-8x^2+1\\ \cos(80^\circ) = 128x^8-256x^6+160x^4-32x^2+1$$ Then $$\sin(10^\circ) = 128x^8-256x^6+160x^4-32x^2+1$$ and using $\sin (a+b) = \sin(a) \cos(b) + \cos a \sin b$ and $\sin(30^\circ)=\frac12$ you get $$\sin(20^\circ)= 256x^9 -512x^7+320x^5-64x^3+2x \\ \cos(50^\circ)=(2x^2-1)(4x^x-3x) - \frac12(256x^9 -512x^7+320x^5-64x^3+2x)\\ \cos(50^\circ) = -128x^9+256x^7-152x^5+22x^3+2x$$ Now that we have all the needed values of cosines, we have to prove that $$(2x^2-1)(4x^x-3x)(8x^4-8x^2+1)=x^2(-128x^9+256x^7-152x^5+22x^3+2x)\\ 64x^9-144x^7+112x^5-34x^3+3x=-128x^{11}+256x^9-152x^7+22x^5+2x^3$$ The tool we have is that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ so $$x^3=\frac{\sqrt{3}/2+3x}{4}$$ We now have to use that replace the highest power of $x$ by 2 powers lower; for example, using $$x^{11} = \frac{3x^9}{4} + \frac{\sqrt{3}}{8}x^8$$ I had to do this replacement trick thirteen times (seven on the right side and six on the left) before I came to $$-x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4} = -x^3 + \frac{\sqrt{3}x^2}{2} + \frac{3x}{4}$$ And we come to the original question -- there has got to be an easier way to show this!

• My first instinct was to use $\cos(x)=\frac{e^{x i}+e^{-x i}}{2}$. It will also involve some brute force, but perhaps it will clean up more rapidly than the work you already did. – JMoravitz Jun 7 '16 at 21:51
• That does not seem to help -- I end up with equally messy expressions for both sides of the equality, and now the trick of knowing and using the value of $\cos\frac{\pi}{6}$ becomes a lot less clean. – Mark Fischler Jun 7 '16 at 21:58

Using degrees as a unit, $$\cos20^{\circ}\cos30^{\circ} = \frac{1}{2}\left(\cos50^{\circ}+\cos10^{\circ}\right)$$ and by multiplying both sides by $\cos40^{\circ}$: $$\cos20^{\circ}\cos30^{\circ}\cos40^{\circ} = \frac{1}{4}\left(\cos90^{\circ}+\cos10^{\circ}+\cos50^{\circ}+\cos30^{\circ}\right).$$ while the same approach leads to: $$\cos10^{\circ}\cos10^{\circ}\cos50^{\circ} = \frac{1}{4}\left(\cos70^{\circ}+2\cos50^{\circ}+\cos30^{\circ}\right)$$ so it is enough to prove that: $$\cos70^{\circ}+\cos50^{\circ}=\cos10^{\circ}$$ that follows from $\cos60^{\circ}=\frac{1}{2}$.

• NIce. This is the sort of answer I was looking for. – Mark Fischler Jun 8 '16 at 5:52
• @Micheal Rozenberg: thank you Micheal, I was too lazy to put a $\circ$ everywhere, but now we may remove the very first line, too. – Jack D'Aurizio Mar 21 '17 at 20:26

Perhaps multiply both sides by $\sin\left(20^{\circ}\right)$, so that \begin{align*} \text{LHS}= & \sin\left(20^{\circ}\right)\cos(20^{\circ})\cos(30^{\circ})\cos(40^{\circ})\\ = & \frac{1}{2}\sin\left(40^{\circ}\right)\cos(30^{\circ})\cos(40^{\circ})\\ = & \frac{1}{4}\sin\left(80^{\circ}\right)\cos(30^{\circ})\\ = & \frac{\sqrt{3}}{8}\cos\left(10^{\circ}\right), \end{align*} and \begin{align*} \text{RHS} & =\left(\sin\left(20^{\circ}\right)\cos(10^{\circ})\cos(50^{\circ})\right)\cos(10^{\circ})\\ & =\frac{1}{2}\sin\left(20^{\circ}\right)\left(\cos(40^{\circ})+\frac{1}{2}\right)\cos(10^{\circ})\\ & =\left(\frac{1}{4}\left(\sin\left(60^{\circ}\right)-\sin(20^{\circ})\right)+\frac{1}{4}\sin\left(20^{\circ}\right)\right)\cos(10^{\circ})\\ & =\frac{\sqrt{3}}{8}\cos\left(10^{\circ}\right). \end{align*}

I may have cheated, but whatever... gonna borrow an identity for a second...

$$\text{if } \alpha+\beta+\gamma = 180 \text{ then } \frac{\sin(2\alpha) + \sin(2\beta) + \sin(2\gamma)} 2 = 2\sin\alpha\sin\beta\sin\gamma$$

$$\cos(20)\cos(30)\cos(40)=\sin(70)\sin(60)\sin(50)$$

Interestingly, this means

$$\cos(20)\cos(30)\cos(40)=\frac{\sin(140)+\sin(120)+\sin(100)}4$$

$$\implies\sin(140)+\sin(120)+\sin(100)=4\cos^2(10)\cos(50)$$

I'm not sure where you'd go from there, possibly try to put the RHS into a form where we can apply the identity?

$$\cos10^\circ=\sin80^\circ=2\sin40^\circ\cos40^\circ=2(2\sin20^\circ\cos20^\circ)\cos40^\circ$$

So, the problem reduces to $$\cos30^\circ=4\cos10^\circ\sin20^\circ\cos50^\circ$$

There is a formal, purely multiplicative proof.

All the information needed is:

• $\cos(20) \cos(40) \cos(80)$ and $\sin(20) \sin(40) \sin(80)$ have known values (in fact we only need the product of $\tan(20) \tan(40)\tan(80)$)
• $\cos (30)$ is known
• basic identities $\sin(x) = \cos(90 - x)$ and $\sin(2x) = 2 \sin x \cos x$

Multiply the alleged formula by $\cos 80 = \sin 10$ , using ( ) to denote known quantities.

c20 c30 c40 c80 ?= s10 c10 c10 c50

(c20 c40 c80) (c30) ?= s10 c10 c10 c50

(c20 c40 c80) (c30) ?= s10 c10 s80 s40

(c20 c40 c80) (c30) ?= (s20 s40 s80) / 2

which is true knowing the values of the (20 40 80) products. The identity is equivalent to the product $\tan(20) \tan(40) \tan(80) = \sqrt{3}$ and does not need separate knowledge of the sine and cosine products.

Until a better reference comes along for the 20-40-80 products, here is a misnamed one, http://en.wikipedia.org/wiki/Morrie%27s_law