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So, I have this equation: $$\sin x = 4\sin10°\sin40°\sin(70°-x)$$ And I'm trying to solve for $x$. Apparently $x=20°$ is the (smallest positive) solution but I can't arrive at it. I'm not very comfortable with this kind of equation so I just try to simplify it using prosthaphaere-something formulas, yet I don't get anywhere but to a messier equation to deal with.

Can someone solve it or at least point me in the right direction? Also, if you could tell me where I can find some material on solving equations like this (I can only find high school trig), that'd be great. Thanks.

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  • $\begingroup$ How many solutions are there on $(0^\circ,70^\circ)$? There must be exactly one using intermediate value concerns. It is clear that the smallest positive solution must be in this interval. $\endgroup$ – Brad Jul 8 '14 at 17:59
  • $\begingroup$ The prostaphaeresis is a little known historical curiosity that was used to perform multiplies before the invention of the logarithms. It is based on product/sum trigonometric identities. en.wikipedia.org/wiki/Prosthaphaeresis $\endgroup$ – Yves Daoust Jul 9 '14 at 7:58
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Using Werner Formulas,

$$2\sin10\sin40=\cos30-\cos50$$

Again, $$4\sin10\sin40\sin(70-x)=2(\cos30-\cos50)\sin(70-x)$$

$$=\sin(100-x)+\sin(40-x)-[\sin(120-x)+\sin(20-x)]$$

Now, $$\sin x=4\sin10\sin40\sin(70-x)$$

$$\iff\sin x=\sin(100-x)+\sin(40-x)-[\sin(120-x)+\sin(20-x)] $$

$$\iff\sin x-\sin(40-x)=[\sin(100-x)-\sin(120-x)]+\sin(x-20) $$

$$\iff2\sin(x-20)\cos20=-2\sin10\cos(110-x)+\sin(x-20)$$

As $\displaystyle\cos(110-x)=\cos\{90-(x-20)\}=\sin(x-20),$ this becomes

$$\sin(x-20)(2\cos20-1)=-2\sin10\sin(x-20)$$

$$\iff\sin(x-20)(2\cos20-1+2\sin10)=0$$

But $\displaystyle\cos20+\sin10=\cos20+\cos80=2\cos50\cos30=\sqrt3\cos50\ne1$

$\displaystyle\implies\sin(x-20)=0\implies x-20=180^\circ n+20$ where $n$ is any integer

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  • $\begingroup$ Thank you! Upvoted good/informative answers but this one serves me best since it doesn't require a calculator or memorizing more identities. (: $\endgroup$ – Deathkamp Drone Jul 10 '14 at 17:07
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Note: All the work here, is in degrees.

First off all it is easy to prove this identity: $\sin(3x)= 4\sin(x)\cdot \sin(60^{\circ}-x) \cdot \sin(60^{\circ}+x)$.

Now, $$ \sin x = 4\sin10°\sin40°\sin(70°-x) $$ $$ \sin x = \frac{4\sin10°\sin50°\sin(70^{\circ})}{\sin(50)\sin(70)} \cdot \sin(40)\cdot \sin(70-x)= \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\sin(70)}$$ $$ \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\sin(70)} = \frac{\sin(30)\sin(40)\sin(70-x)}{\sin(50)\cos(20)} = \frac{\sin(20)\sin(70-x)}{\sin(50)} $$

So we are left with, $\sin(x) \sin(50) = sin(20) \sin(70-x)$ . Its clear that $x=20$.One can prove it is only solution, by expanding ..

$$\sin(x)\sin(50) = \sin(20)\sin(70)\cos(x)-\sin(20)\sin(x)\cos(70)$$ $$ \sin(70) \cos(20) - \sin(20)\cos(70) = \sin(70-20)= \sin(20)\sin(70)\cot(x)-\sin(20)\sin(70) $$ Thus $ \tan(x) = \tan(20) $

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    $\begingroup$ Mind sharing where the first identity came from? $\endgroup$ – Deathkamp Drone Jul 8 '14 at 19:04
  • $\begingroup$ Sorry, earlier there was some typo , now its right :) . Proof: $$4\sin(x)(\sin(60)\cos(x)-\sin(x)\cos(60))(\sin(60)\cos(x)-\sin(x)\cos(60)) = 4\sin(x)(\sin^2(60)\cos^2(x)-\sin^2(x)\cos^(60)) = \sin(x)(3\cos^2(x)-\sin^2(x)) = 3\sin(x) - 4 \sin^3(x) = \sin(3x)$$ $\endgroup$ – Shivang jindal Jul 9 '14 at 3:23
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    $\begingroup$ @DeathkampDrone, cut-the-knot.org/arithmetic/algebra/sin3x.shtml $\endgroup$ – lab bhattacharjee Jul 9 '14 at 6:28
  • $\begingroup$ And how can we, poor mortals, make the connection between this lemma and the starting equation ? $\endgroup$ – Yves Daoust Jul 9 '14 at 12:50
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A way to address such trigonometric equations is to express everything in terms of the sine and cosine of the unknown. In your case, you can get rid of the argument $70°-x$ using the angle subtraction formula.

So, develop $$\sin x=4\sin10°\sin40°\sin70°\cos x-4\sin10°\sin40°\cos70°\sin x.$$ Doing that, you obtain a linear trigonometric equation of the form $$A\cos x+B\sin x=C,$$ for which a general solution is known.

But your case is simpler, as $C=0$, and you directly derive $$\tan x=\frac{4\sin10°\sin40°\sin70°}{1+4\sin10°\sin40°\cos70°}.$$ Using a calculator, you find $x=20°$.

As $\sin x=0$ is not a solution, you didn't change the equation by this transformation. And given that the tangent function has a half turn period, the complete solution is $$x=20°+180°k.$$

Now you easily prove that $x=20°$ is an exact solution with $$4\sin10°(\sin40°\sin50°)=4\sin10°\frac12(\cos10°-\cos90°)=2\sin10°\cos10°=2\frac12(\sin20°-\sin0°)=\sin20°,$$ by two applications of the the prosthaphaeresis formulas.

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Hint: Use the fact that $\sin{x-y}=\sin x \cos y-\cos x \sin y$ to expand $\sin(70°-x)$. From there you should be able to solve for $x$.

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  • $\begingroup$ Sorry, but I fail to see how it helps. I've tried doing that before, just to get to $4\sin10°\sin40°(\sin70°\mathrm{cot}\,x + \cos70°) = 1$. It's nice because now we only have one $x$ to solve for but I'm still clueless as for how to solve for $x$ from there. $\endgroup$ – Deathkamp Drone Jul 8 '14 at 18:14
  • $\begingroup$ @DeathkampDrone you would then rearrange to get $\cot x$ on one side, then take the inverse cotangent. $\endgroup$ – zeta Jul 8 '14 at 19:06
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}\newcommand{\c}{^{\circ}}$ $\ds{\sin\pars{x} = 4\sin\pars{10\c}\sin\pars{40\c}\sin\pars{70\c - x}:\ {\large ?}}$

\begin{align} &4\sin\pars{10\c}\sin\pars{40\c} ={\sin\pars{35\c}\cos\pars{35\c - x} - \cos\pars{35\c}\sin\pars{35\c - x}\over \sin\pars{35\c}\cos\pars{35\c - x} + \cos\pars{35\c}\sin\pars{35\c - x}} \\[3mm]&={1 - \cot\pars{35\c}\tan\pars{35\c - x}\over 1 + \cot\pars{35\c}\tan\pars{35\c - x}} \\[3mm]&\imp\quad{4\sin\pars{10\c}\sin\pars{40\c} + 1 \over 4\sin\pars{10\c}\sin\pars{40\c} - 1} ={2 \over -2\cot\pars{35\c}\tan\pars{35\c - x}} \end{align}

$$ \tan\pars{35\c - x}= {1 - 4\sin\pars{10\c}\sin\pars{40\c} \over 1 + 4\sin\pars{10\c}\sin\pars{40\c}}\, \tan\pars{35\c} $$

$$ x = 35\c -\ \underbrace{\arctan\pars{{1 - 4\sin\pars{10\c}\sin\pars{40\c} \over 1 + 4\sin\pars{10\c}\sin\pars{40\c}}\, \tan\pars{35\c} }}_{\ds{15\c - n\times 180\c\,,\quad n \in {\mathbb Z}}} $$

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