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Prove that:

$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$

My Approach:

$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$ $$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$ $$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$

Now, please help me to continue from here.

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    $\begingroup$ I would expand $\cos(120^\circ + A)$ and $\cos(240^\circ + A)$. $\endgroup$ – Arthur Sep 10 '16 at 6:16
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There's also another more algebraic way. You can easily show (by expansion) that if $$a+b+c=0$$ then $$a^3+b^3+c^3=3abc$$ Since, in your problem, for every $A$ $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ Then you can use the above identity $$\cos^3{A}+\cos^3{(A+2\pi/3)}+\cos^3{(A-2\pi/3)}=3\cos{A}\cos{(A+2\pi/3)}\cos{(A-2\pi/3)}$$ By using the previously mentioned identity $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos{(x+y)}+\cos{(x-y)}\right)$$ you can simplify the RHS to this $$\frac{3}{2}\cos{A}\left(\cos{(2A)}-\frac{1}{2}\right)$$ and then (using it again) $$\frac{3}{4}\left(\cos{3A}+\cos{(A)}\right)-\frac{3}{4}\cos{A}=\frac{3}{4}\cos{(3A)}$$

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  • $\begingroup$ How is $a+b+c=0$ $\endgroup$ – pi-π Sep 10 '16 at 7:49
  • $\begingroup$ This is a well-known identity. Just expand $\cos{(A+2\pi/3)}$ and $\cos{(A-2\pi/3)}$ (by using $\cos(\alpha+\beta)$ formula) and then add $\cos{A}$ to see the terms cancel out each other. $\endgroup$ – Babak Sep 10 '16 at 7:58
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You can use this particular formula: $$\cos{x}\cos{y}=\frac{1}{2}\left(\cos(x+y)+\cos(x-y)\right)$$ twice and simplify $\cos^3{A}$, $\cos^3{(A+2\pi/3)}$, and $\cos^3{(A-2\pi/3)}$ in this way: $$\begin{aligned} \cos^3{A}&=\cos{A}\left(\cos{A}\cos{A}\right) \\ &=\frac{1}{2}\cos{A}\left(1+\cos{2A} \right) \\ &=\frac{1}{2}\cos{A}+\frac{1}{2}\cos{A}\cos{2A}\\ &=\frac{1}{2}\cos{A}+\frac{1}{4}\left(\cos{A}+\cos{3A}\right)\\ &=\frac{1}{2}\cos{A}+\frac{1}{4}\cos{A}+\frac{1}{4}\cos{3A}\\ \end{aligned}$$

And

$$\begin{aligned} \cos^3{(A+2\pi/3)}&=\frac{1}{2}\cos{(A+2\pi/3)}\left(1+\cos{(2A+4\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A+2\pi/3)}\left(1-\cos{(2A+\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{2}\cos{(A+2\pi/3)}\cos{(2A+\pi/3)}\\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\left(\cos{(A-\pi/3)}-\cos{3A}\right)\\ &=\frac{1}{2}\cos{(A+2\pi/3)}-\frac{1}{4}\cos{(A-\pi/3)}+\frac{1}{4}\cos{3A} \end{aligned}$$

And

$$\begin{aligned} \cos^3{(A-2\pi/3)}&=\frac{1}{2}\cos{(A-2\pi/3)}\left(1+\cos{(2A-4\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A-2\pi/3)}\left(1-\cos{(2A-\pi/3)} \right) \\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{2}\cos{(A-2\pi/3)}\cos{(2A-\pi/3)}\\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\left(\cos{(A+\pi/3)}-\cos{3A}\right)\\ &=\frac{1}{2}\cos{(A-2\pi/3)}-\frac{1}{4}\cos{(A+\pi/3)}+\frac{1}{4}\cos{3A} \end{aligned}$$

Now, by adding the results and using the following identities you can get your answer (the second one can be proved by using the first identity in the reverse order). $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ $$\cos{(A-\pi/3)}+\cos{(A+\pi/3)}=\cos{A}$$

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Consider the following picture. We see that the triangle with vertices $(\cos A, \sin A), (\cos(A+120^\circ), \sin(A+120^\circ)), (\cos(A+240^\circ), \sin(A+240^\circ))$ are the vertices of an equilateral triangle with centriod at the origin. enter image description here

Thus if we set $a = \cos A, b = \cos(A+120^\circ), c = \cos(A+240^\circ)$, then \begin{align*} a+b+c = 0 \end{align*} and hence \begin{align*} a^3+b^3+c^3 = 3abc \end{align*} and \begin{align*} \cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)} &= 3\cos A \cos(A+120^\circ)\cos(A+240^\circ) \\ &= \frac{3}{2}(2\cos A \cos(A+120^\circ))\cos(A+240^\circ) \\ &= \frac{3}{2}[\cos(2A+120^\circ)+\cos(120^\circ)]\cos(A+240^\circ)\\ &=\frac{3}{2}\cos(2A+120^\circ)\cos(A+240^\circ)-\frac{3}{4}\cos(A+240^\circ)\\ &=\frac{3}{4}\cos(3A+360^\circ)+\frac{3}{4}\cos(A-120^\circ) -\frac{3}{4}\cos(A+240^\circ)\\ &=\frac{3}{4}\cos(3A) \end{align*} since $\cos(A+240^\circ) = \cos(360^\circ - (A+240^\circ)) = \cos(120^\circ -A) = \cos(A-120^\circ)$

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  • $\begingroup$ If you can see geometrically, the points $(\cos A, \sin A), (\cos(120^\circ+A), \sin(120^\circ+A)), (\cos(240^\circ), \sin(240^\circ+A))$ are the vertices of an equilateral triangle with centroid as the origin, then the above proof holds since we have used only an algebraic identity. $\endgroup$ – user348749 Sep 10 '16 at 7:23
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use that $$\cos(120^{\circ}+x)=-1/2\,\cos \left( x \right) -1/2\,\sqrt {3}\sin \left( x \right) $$ and $$\cos(240^{\circ}+x)=-1/2\,\cos \left( x \right) +1/2\,\sqrt {3}\sin \left( x \right) $$

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For the LHS, use $\cos(120^\circ +A)=\cos 120^\circ \cos A - \sin 120^\circ \sin A$

and $\cos(240^\circ+A)=\cos 240^\circ \cos A - \sin 240^\circ \sin A$

Then $\cos^3(120^\circ+A)=\left(\cos 120^\circ \cos A - \sin 120^\circ \sin A \right)^3$

and $\cos^3(240^\circ+A)=\left( \cos 240^\circ \cos A - \sin 240^\circ \sin A \right)^3$

Pop in values of $\cos 120^\circ$, $\sin 120^\circ$, etc before expanding brackets.

$\cos^3(120^\circ+A)=\left(-1/2 \cos A -\sqrt 3/2 \sin A \right)^3=-\frac 18 \left(\cos A +\sqrt 3 \sin A \right)^3$

and $\cos^3(240^\circ+A)=-\frac 18\left( \cos A - \sqrt3\sin A \right)^3$

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  • $\begingroup$ It became quiet complex. Could you please suggest some convenient method? $\endgroup$ – pi-π Sep 10 '16 at 6:30
  • $\begingroup$ Unless you want to use complex numbers (see other answer) you just have to persevere. Do you know how to use binomial expansions? $\endgroup$ – tomi Sep 10 '16 at 6:48
  • $\begingroup$ yes. I know about that. $\endgroup$ – pi-π Sep 10 '16 at 6:53
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Hint: As mentioned in the comments, substitute: $cos(120+A)$ and $sin(120+A)$ :$$\color{blue}{cos(120+A)=\dfrac{-1}{2}cosA-\dfrac{\sqrt{3}}{2}sinA}$$ $$\color{red}{cos(240+A)=\dfrac{-1}{2}cosA+\dfrac{\sqrt{3}}{2}sinA}$$

The L.H.S becomes: $${cos}^3A +(\color{blue}{\dfrac{-1}{2}cosA-\dfrac{\sqrt{3}}{2}sinA})^3+(\color{red}{\dfrac{-1}{2}cosA+\dfrac{\sqrt{3}}{2}sinA})^3$$ Simplifying this further and using the fact that $\color{blue}{a}^3+\color{red}{b}^3=(\color{blue}{a}+\color{red}{b})(\color{blue}{a}^2-\color{blue}{a}\color{red}{b}+\color{red}{b}^2)$ we get: $${cos}^3A-\dfrac{1}{4}{cos}^3A-\dfrac{9}{4}{sin}^2AcosA=\color{green}{\dfrac{3}{4}({cos}^3A+3{sin}^2AcosA)}$$

Note that the R.H.S is equal to: $$\color{green}{\dfrac{3}{4}({cos}^3A+3{sin}^2AcosA)}$$

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First note that $\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta$. Observing that $$\cos(3A+3\cdot\tfrac23\pi)=\cos(3A+3\cdot\tfrac43\pi)=\cos3A,$$and adding the results of $\cos^3\theta$ for $\theta=A$, $\theta=A+\frac23\pi$, and $\theta=A+\frac43\pi$, using the fact that $\cos A+\cos(A+\frac43\pi)=2\cos(A+\frac23\pi)\cos\frac23\pi=-\cos(A+\frac23\pi)$, now gives the required form.

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If $\cos3x=\cos3A$

$3x=n360^\circ\pm3A$ where $n$ is any integer

$x=n120^\circ+ A$ where $n=0,1,2$

As $\cos3x=4\cos^3x-3\cos x,$ the roots of $4c^3-3c-\cos3A=0$ are

$c_{n+1}=\cos(n120^\circ+ A)$ where $n=0,1,2$

Using Vieta's formula, $$c_1+c_2+c_3=0, c_1c_2c_3=\dfrac{\cos3A}4$$ $$\text{ and }c_1c_2+c_2c_3+c_3c_1=-\dfrac34$$

We need $c_1^3+c_2^3+c_3^3$

Now use $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

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