I have read an article in Wikipedia about solving of systems of polynomial equations, where for example if we consider equation using trigonometric function,we should change $\sin(x)$ and $\cos(x)$ with variables $s$ and $c$ and add new equation $s^2+c^2-1=0$; consider this


The page says that this is equivalent to this

$$\begin{align} s^3+4c^3-3c&=0\\ s^2+c^2-1&=0 \end{align}$$

I don't understand why, could you explain to me why are they same?

  • $\begingroup$ The way to make trig identities much simpler is to just use Euler's identity; then you only have simple ratios of sums of exponential functions. I gave up memorizing trig identities long ago and just look at a table or use Euler if one isn't handy/helpful. $\endgroup$ – William Grobman Oct 8 '11 at 18:24

The identity $\cos 3x=4\cos^3x-3\cos x$ was used. That transforms the given equation into the first of the pair of equations. The second equation of the pair is just the one that you had in the text. However, in both cases it should be $c^2$, not $c*2$.

The identity for $\cos3x$ can be derived like this:

$$ \begin{align} \cos^3x &=\left(\frac12\left(\mathrm e^{\mathrm ix}+\mathrm e^{-\mathrm ix}\right)\right)^3\\ &=\frac18\left(\mathrm e^{3\mathrm ix}+3\mathrm e^{\mathrm ix}+3\mathrm e^{-\mathrm ix}+\mathrm e^{-3\mathrm ix}\right)\\ &=\frac14\cos3x+\frac34\cos x\;. \end{align} $$

  • $\begingroup$ aaa yes ,just typo because of keyboard and shift key,thanks @joriki $\endgroup$ – dato datuashvili Oct 8 '11 at 16:36

What simply happened here is that you first want everything in terms of sines and cosines without multiple arguments. Since $\cos\,3x=4\cos^3\,x-3\cos\,x$, making that substitution into your original equation yields $\sin^3 x+4\cos^3\,x-3\cos\,x=0$.

Now, morally this is an equation in a single variable. However, one could skip having to tangle with transcendentals if we let $c=\cos\,x$ and $\sin\,x$, yielding $s^3+4c^3-3c=0$. But in going to this form, we now have two variables to contend with. To undo the extra degree of freedom we added, we now add the additional equation $c^2+s^2=1$, which is from the Pythagorean formula. Now you have two equations in two unknowns...

An alternative is to consider the Weierstrass substitution. If we let $c=\dfrac{1-t^2}{1+t^2}$ and $s=\dfrac{t}{1+t^2}$, we again end up with a single equation in a single unknown, which can be solved by purely algebraic means. As soon as you obtain values of $t$, one can retrieve the corresponding $x$, since $x=\tan\dfrac{t}{2}$... (why?)

  • $\begingroup$ thanks @J.M ,i am happy because studed what did not know before $\endgroup$ – dato datuashvili Oct 8 '11 at 16:45

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