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Aside from the obvious knowledge that the roots of $\sin x$ are all integer multiples of $\pi$, is there a formal, algebraic method to calculate the roots of trigonometric functions similar to the quadratic equation?

(e.g. roots of $\sin^2(ax) + \sin(bx) + c$ or some other non-trivial form)

Amzoti
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Jason Nichols
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1 Answers1

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While trigonometric identities exist for multiples of an angle, perhaps the best-known being:

$$ \begin{align} \sin 2x &= 2\sin x \cos x\\ \cos 2x &= \cos^2 x - \sin^2{x} =2\cos^2 x - 1 = 1 - 2 \sin^2x\\ \sin 3x &= 3\sin x - 4 \sin^3 x\\ \cos 3x &= 4\cos^3 x - 3\cos x \end{align} $$

there is no such general formula, and indeed no way to solve such equations, unless the trigonometric arguments are the same and therefore the expression can be factored.

Lee Sleek
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    using those identities, is there a simpler root for something like $\cos ax + \cos bx$ ? – Jason Nichols Jun 11 '13 at 03:54
  • You would have to use the individual cosine identities for a and b to expand the expressions and then solve. For example, try using the identities I just gave to solve $\cos 3x - 2\cos 2x = 0$. – Lee Sleek Jun 11 '13 at 03:58
  • I follow your expansion to $4\cos^3 x - 3 \cos x - 4 \cos^2 x =2 $ but without resorting to the trivial integer multiple of $(n+(1/2) ) \pi $, I'm not with you. – Jason Nichols Jun 11 '13 at 04:08
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    @JasonNichols for $\cos(ax)+\cos(bx)=0$ use sum to product identities. But in general there is no method for solving a generic $f(x)=0$ other than numerical approximation. – Maesumi Jun 11 '13 at 04:15
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    @JasonNichols Try using the cubic formula. – Lee Sleek Jun 11 '13 at 14:29
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    @JasonNichols, for $\cos ax + \cos bx$ you could use the identity $$\cos ax + \cos bx = 2 \cos\left(\frac{a+b}{2},x\right) \cos\left(\frac{a-b}{2},x\right)$$ then find the zeros of each cosine in the product. – Antonio Vargas Jun 12 '13 at 12:35