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If $x_1, x_2$ are roots of $a\cos x+b\sin x+c=0$ for $x_1+x_2≠2kπ$ show that $\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$

What I've done till now:

$$a\cos x_1+b\sin x+c-(a\cos x_2+b\sin x_2+c)=0$$

$a\cos x_1+b\sin x_1-a\cos x_2-b\sin x_2=0$

$a(\cos x_1-\cos x_2)+b(\sin x_1-\sin x_2)=0$

$a(-2\sin(\frac{x_1+x_2}{2})\sin(\frac{x_1-x_2}{2}))+b(2\sin(\frac{x_1-x_2}{2})\cos(\frac{x_1+x_2}{2}))=0$

$-a\sin(\frac{x_1+x_2}{2})+b(\frac{\cos x_1+x_2}{2})=0$

That's it. Does anyone know how to solve this?

mathstudent1
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EL02
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    Check this: https://math.stackexchange.com/q/927530/42969 – Martin R Jan 17 '21 at 21:44
  • Use $\sin x$ for $\sin x$. –  Jan 17 '21 at 21:44
  • Your way of using MathJax is abominable, but I've cleaned up part of it. Maybe you can use that to figure out how clean up the rest. – Michael Hardy Jan 17 '21 at 21:51
  • See https://math.stackexchange.com/questions/3989321/if-x-1-x-2-are-roots-of-a-cos-xb-sin-xc-0-for-x-1x-2-ne-2k-pi-show-tha, and https://math.stackexchange.com/questions/2753518/if-alpha-and-beta-are-distinct-roots-of-the-equation-p-cos-theta-q – lab bhattacharjee Jan 18 '21 at 05:21

1 Answers1

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Just rewrite this equation in terms of complex exponentials:

$$\frac{a-bi}{2}e^{ix}+\frac{a+bi}{2}e^{-ix}+c=0$$

Setting $z=a+bi$ one can rewrite the equation as a quadratic in $w=e^{ix}$ $$\bar{z}w^2+cw+z=0$$ From Vieta's formulae we know that the product of the two roots of the equation is

$$w_1w_2=e^{i(x_1+x_2)}=\frac{z}{\bar{z}}$$

which, upon separating real and imaginary parts yields

$$\cos(x_1+x_2)=\frac{a^2-b^2}{a^2+b^2}~,~\sin(x_1+x_2)=\frac{2ab}{a^2+b^2}$$

DinosaurEgg
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  • thanks for the answer, but I'll take the algebra class next semester and still haven't learnt complex numbers and such. – EL02 Jan 17 '21 at 22:20