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Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$

My Attempt:

$$L.H.S=\cos^2 20° + \cos^2 40° + \cos^2 80°$$ $$=\dfrac {1+\cos 40}{2}+\dfrac {1+\cos 80}{2} + \dfrac {1+\cos 160°}{2}$$ $$=\dfrac {3+\cos 40°+\cos 80°+\cos 160°}{2}$$

I.could not solve further from here..

pi-π
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  • More generally, $$\cos^2 n° + \cos^2 (60-n)° +\cos^2 (60+n)° = \sin^2 n° + \sin^2 (60-n)° +\sin^2 (60+n)° $$ where the post had $n=20$. As well as, $$\cos^2 n° + \cos^2 45° +\cos^2 (90-n)° = \sin^2 n° + \sin^2 45° +\sin^2 (90-n)° $$ – Tito Piezas III Feb 08 '17 at 15:22

2 Answers2

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We know the identity $$\cos^2 x - \sin^2 x = \cos 2x \tag {1}$$ and $$\cos A - \cos B = 2\cos \frac {A+B}{2} \cos \frac {A-B}{2} \tag {2} $$

We have $$(\cos^2 20^\circ - \sin^2 20^\circ) + (\cos^2 40^\circ - \sin^2 40^\circ) + (\cos^2 80^\circ - \sin^2 80^\circ) $$ $$= \cos 40^\circ + \cos 80^ \circ + \cos 160^ \circ $$ (using $(1)$) $$ =2\cos 60^\circ \cos 20^\circ +\cos (90^\circ + 70^\circ) $$ (using $(2)$) $$ = \cos 20^\circ - \sin 70^\circ =0$$ where we used the identity $\cos (90^\circ + x) = -\sin x $.

Hope it helps.

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I shall prove the generalization of $$\cos x+\cos(x+120^\circ)+\cos(x+240^\circ)=0$$ (here $x=40^\circ$)

$$S=\sum_{r=0}^{n-1}\cos\left(x+\dfrac{360^\circ r}n\right)=0$$

Method $\#1:$

$S=$ real part of $\sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r}n\right)$

Now $\displaystyle \sum_{r=0}^{n-1} \exp i\left(x+\dfrac{360^\circ r} n \right) = \frac{e^{i360^\circ}-1}{\exp i\left(x+\dfrac{360^\circ(-n)}n\right)-1}=0$

Now equate the real & the imaginary parts.

Method $\#2:$ Using multiple angle formula cosine:

$$\cos(nx)=2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots$$

Now if $\cos nx=\cos ny, nx=360^\circ m\pm ny$ where $m$ is any integer

$x= y+\dfrac{360^\circ m}n$ where $m\equiv0,1,2,\ldots, n-1\pmod n$

So, the roots of $$2^{n-1}\cos^nx-n2^{n-3}\cos^{n-2}x+\cdots-\cos ny=0$$ are

$\cos\left(y+\dfrac{360^\circ m}n\right)$ where $m\equiv0,1,2,\ldots n-1\pmod n$

Using Vieta's formula,

$$\sum_{r=0}^{n-1}\cos\left(y+\dfrac{360^\circ m}n\right)=\dfrac0{2^{n-1}}$$

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lab bhattacharjee
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  • @ lab bhattacharjee, what is Vieta's formula? – pi-π Feb 08 '17 at 18:04
  • @Euler, https://www.google.co.in/search?q=Vieta%27s+formula – lab bhattacharjee Feb 09 '17 at 07:04
  • @labbhattacharjee : The proper way to code $a\exp b$ is a\exp b. That gets you proper spacing in in things like $a\exp b$ and $a\exp(b)$, and I include both of those examples so that you can see that the context-dependent nature of the spacing is handled by the software. – Michael Hardy May 10 '17 at 16:51