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This is a part of a larger question.

I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:

$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \sqrt{3} \sin \alpha)$$

I have to use the above and the following results: $\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2\sqrt{2}}$ to find the solutions of the following:

$$y^3-3y-\sqrt{2}=0$$

I assumed that the constant term must be the equivalent of the cosine term and tried to find alpha so that I have one solution and then can derive the other. But since $\arccos {\sqrt{2}}$ is not trivially defined, this is not a correct approach. Or at least it is not correct the way I am doing it. Also I would have to do a bit more trigs since the second polynomial is not equivalent to the first one. There must be an easier, neater solution.

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  • $\begingroup$ Are you familiar with the concept of a Depressed Cubic Equation? en.wikipedia.org/wiki/… $\endgroup$
    – Xoque55
    Dec 6, 2015 at 16:18
  • $\begingroup$ Notice, to apply above $4x^3$ thing, you have to get {4,0,-3} as first 3 coefficients. $y^3$ is not $4y^3$. Try a substitution like $y=kx$ so that you get those coefficients. $\endgroup$ Dec 6, 2015 at 16:18

5 Answers 5

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Using your hints: If $y$ satisfies $y^3-3y-\sqrt{2}=0$, try setting $y=kx$ for $k$ to be chosen later. Then $x$ satisfies $$k^3x^3-3kx-\sqrt2=0\tag1$$ If $k=2$, the LHS of (1) becomes $$ 8x^3-6x-\sqrt2=2\left(4x^3-3x-\frac{\sqrt2}2\right), $$ which means that $x$ satisfies the famous equation $$ 4x^3-3x-\cos 3\alpha=0\tag{*} $$ with $\alpha=15^\circ$. Now apply the factorization of (*) you've been given and your two results about $\sin 15^\circ$ and $\cos 15^\circ$ to obtain the three possible solutions for $x$. Finally, get the solutions of the original equation knowing that $y=kx=2x$.

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Recall the triplication formula: $$ \cos3\alpha=4\cos^3\alpha-3\cos\alpha $$ so, if your equation is $y^3-3y=\sqrt{2}$, you can first set $y=at$, so $$ a^3t^3-3at=\sqrt{2} $$ and you'd like that $a^3/3a=4/3$, so you can take $a=2$: $8t^3-6t=\sqrt{2}$; setting $t=\cos\alpha$, we get $$ \cos3\alpha=\frac{\sqrt{2}}{2} $$ so $$ 3\alpha=\frac{\pi}{4}+2k\pi $$ where $k=0$, $k=1$ or $k=2$. Hence $$ \frac{\pi}{12},\quad\frac{3\pi}{4},\quad\frac{17\pi}{12} $$ are the solutions for $\alpha$.

Now, $$ 2\cos\frac{\pi}{12}=2\sqrt{\frac{1+\cos(\pi/6)}{2}}= \sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2} $$ then $$ 2\cos\frac{3\pi}{4}=-\sqrt{2} $$ and $$ 2\cos\frac{17\pi}{12}=-2\cos\frac{5\pi}{12}= -2\sqrt{\frac{1+\cos(5\pi/6)}{2}}= -\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2} $$

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If $x=\cos A,$ using $\cos3y=4\cos^3y-3\cos y$

$$\cos3A=\cos3\alpha\implies3A=2n\pi\pm3\alpha$$ where $n$ is any integer

$$\implies A=\dfrac{2n\pi}3+\alpha$$ where $n\equiv0,1,2\pmod3$


Alternatively, $$4x^3-3x-(4\cos^3\alpha-3\cos\alpha)=0$$

$$4(x^3-\cos^3\alpha)-3(x-\cos\alpha)=0$$

Take out $$x-\cos\alpha$$ as common factor & solve the remaining quadratic equation of $x$

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Thee trigonometric method relies on the trigonometric identity: $$\cos 3\theta=4\cos^3\theta -3\cos\theta.$$ Set $\;y=A\cos\theta,\enspace A>0$. The equation becomes $$A^3\cos^3\theta -3A\cos\theta=\sqrt2$$ We choose $A>0$ such that the left-hand side of the equation is proportional to the development of $\cos3\theta $, i.e. $$\frac{A^3}4=\frac{3A}3\iff A^2=4.$$ Thus we set $y=2\cos\theta $, so that we get the trigonometric equation $$2\cos3\theta =\sqrt2\iff \cos3\theta =\frac{\sqrt2}2=\cos\frac\pi4,$$ and the solutions are $$3\theta \equiv\pm\frac\pi4\mod2\pi\iff\theta \equiv\pm\frac\pi{12}\mod\frac{2\pi}3$$ There are $3$ different values for $\cos\theta$, and the different values for $y$ are $$2\cos\frac\pi{12},\quad2\cos\frac{3\pi}4=-\sqrt2,\quad2\cos\frac{-7\pi}{12}=-\sin\frac\pi{12}$$ Now, the addition formulae yield: \begin{align*} \cos\frac\pi{12}&=\cos\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6+\sqrt2}4,\\ \sin\frac\pi{12}&=\sin\Bigl(\frac\pi3-\frac\pi4\Bigr)=\dots=\frac{\sqrt6-\sqrt2}4. \end{align*}

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To convert the second equation into the form of the first equation, let $y=2x$ and divide out by $2$ to obtain:

$$4x^3-3x-\frac{\sqrt{2}}{2}=0$$

Thus, the first solution is $\cos 3\alpha = \frac{\sqrt{2}}{2}$ or $\alpha = \pm\frac{\pi}{12}+2n\pi$, $x=\cos{\frac{\pi}{12}}$ or $y=2\cos{\pi/12}$. The other solutions are easily found applying the factorization obtained in the first equation.

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